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http://www.patricioginelsa.com/lib/introduction-to-statistics-and-data-analysis-student-study-guide-and-workbook
|
# Introduction to Statistics and Data Analysis Student Study
Format: Paperback
Language: English
Format: PDF / Kindle / ePub
Size: 5.38 MB
The hypotheses are H0: ρ = 0 H1: ρ = 0 Step 2: Calculate the Critical Value For a two-tailed test using ν = n − 2 = 30 − 2 = 28 degrees of freedom.8 Scatter plots for 30 MBA applicants 30 Randomly Chosen MBA Applicants 4.05) and for the two-tailed p-value for GMAT score.00 4. the rejection is not very compelling.0161 (reject ρ = 0 since p < .862 > tα = 2. 2 .4356) between verbal and quantitative GMAT scores statistically significant? Fitzmaurice and Geert Molenberghs Goodness-of-fit tests for ordinal response regression models. .. .. .. 175--190 Peter W.
Pages: 300
Publisher: Worth Publishers; 716th edition (December 15, 1997)
ISBN: 0716731975
Study Guide to Accompany Fundamental Statistics for Behavioral Sciences
Weighted Least Squares Regression, 2013 Edition (Statistical Associates Publishers Blue Book Series 43)
Chambers Design-Adjusted Regression with Selectivity Bias. .. .. .. .. .. . 323--334 Sander Greenland and Ruth M. Mickey Closed form and dually consistent methods for inference on strict collapsibility in $2\times 2\times K$ and $2\times J\times K$ tables. .. .. 335--343 Tak K Non-Obvious: How to Think download online http://makinabezi.com/freebooks/non-obvious-how-to-think-different-curate-ideas-predict-the-future. The expertly designed curriculum enables you to use industry-standard software such as Minitab, R, and SAS to improve your data analysis proficiency. In two to five years you can complete the degree, selecting from courses covering a variety of statistical applications areas, including: Students who successfully complete the master's program have the option to prepare for the SAS Base Programming Certification Exam, or to seek PStat® accreditation through the American Statistical Association as an Accredited Professional StatisticianTM Mdr's School Directory Utah 2005-2006 read for free. Famoye F., Goodness of fit tests for generalized logarithmic series distribution, Computational Statistics and Data Analysis, 33, 59-67, 2000. Fan Y., Goodness-of-fit tests for a multivariate distribution by the empirical characteristic function, Journal of multivariate analysis, 62, 1997, 36 , e.g. Statistical Methods For read epub http://www.patricioginelsa.com/lib/statistical-methods-for-communication-science. In the Hellenic experience this kind of wisdom received a more structural character in the form of philosophy , e.g. Frontiers in Data Science (Chapman & Hall/Crc Big Data) read for free. Alternatively, you can paraphrase another person’s work, but be sure you are not just rearranging or replacing a few words. A good strategy is to (1) read over what you want to paraphrase carefully, (2) cover up the text with your hand, (3) write out the idea in your own words without peeking, and (4) check your paraphrase against the original text to be sure you have not accidentally used the same phrases or words, and that the information is accurate. 3 , source: Homework Manager Plus Passcode read online c4bmediawebsites.com. Suppose you hear an "old-timer" say, "Why, in my day, kids were much more respectful and didn't cause as much trouble as they do nowadays!" Formulate a hypothesis related to this statement that you could test. The calculation of a p-value is based on the assumption that a finding is the product of chance alone, given the parameters of a level of significance (for example, choosing a level of significance at 0.05, if your p-value is found to be at 0.05 or lower, then you can conclude with 5% or lower confidence that the results from your experiment were purely due to chance) Mathematical Statistics drrajaratnam.com.
Mississippi in Perspective 2006
Study Guide for Anderson/Sweeney/Williams' Quantitative Methods for Business, 10th
CengageNOW on WebCT(TM) Instant Access Code for Anderson/Sweeney/Williams' Essentials of Modern Business Statistics
Introduction to Statistics in Psychology
Working With Numbers and Statistics: A Handbook for Journalists (Routledge Communication Series)
Data Analysis with Mplus (Methodology in the Social Sciences)
Pennsylvania Crime in Perspective 2004
Mdr's School Directory Illinois 2006-2007: Spiral Edition
Introduction to Statistical Analysis
Business StatisticsNOW(TM) Online Access for Anderson/Sweeney/Williams' Statistics for Business and Economics, 9th
Naturalistic Evaluation (Jossey Bass Social and Behavioral Science Series)
Bundle: Statistical Methods for Engineers, 3rd + MINITAB Student Version 14 for Windows
Statistical Thinking for Managers (Duxbury Series in Statistics and Decision Sciences)
The Improbability Principle: Why Coincidences, Miracles, and Rare Events Happen Every Day
Vermont in Perspective 2006
County Business Patterns New Jersey 2000
SPSS 7 5 for Windows Brief Guide
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2017-08-20 05:46:46
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https://proofwiki.org/wiki/Z/(m)-Module_Associated_with_Ring_of_Characteristic_m
|
# Z/(m)-Module Associated with Ring of Characteristic m
## Theorem
Let $\left({R,+,*}\right)$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let the characteristic of $R$ be $m$.
Let $\left({\Z_m, +_m, \times_m}\right)$ be the ring of integers modulo $m$.
Let $\circ$ be the mapping from $\Z_m \times R$ to $R$ defined as:
$\forall \left[\!\left[a\right]\!\right]_m \in \Z_m: \forall x \in R: \left[\!\left[a\right]\!\right]_m \circ x = a \cdot x$
where $\left[\!\left[a\right]\!\right]_m$ is the residue class of $a$ modulo $m$ and $a \cdot x$ is the $a$th power of $x$.
Then $\left({R, +, \circ}\right)_{\Z_m}$ is a unitary $\Z_m$-module.
## Proof
Let us verify that the definition of $\circ$ is well-defined.
Let $\left[\!\left[a\right]\!\right]_m=\left[\!\left[b\right]\!\right]_m$.
Then we need to show that:
$\forall x \in R : \left[\!\left[a\right]\!\right]_m \circ x = \left[\!\left[b\right]\!\right]_m \circ x$
By the definition of congruence:
$\left[\!\left[a\right]\!\right]_m = \left[\!\left[b\right]\!\right]_m \iff \exists k \in \Z : a = b + k m$
Then:
$\displaystyle \left[\!\left[a\right]\!\right]_m \circ x$ $=$ $\displaystyle a \cdot x$ Definition of $\circ$ $\displaystyle$ $=$ $\displaystyle \left(b+km\right) \cdot x$ $\displaystyle$ $=$ $\displaystyle b \cdot x + km \cdot x$ Powers of Group Elements: Sum of Indices $\displaystyle$ $=$ $\displaystyle b \cdot x + k \cdot \left( m \cdot x \right)$ Powers of Group Elements: Product of Indices $\displaystyle$ $=$ $\displaystyle b \cdot x + k \cdot 0_R$ Characteristic times Ring Element is Ring Zero $\displaystyle$ $=$ $\displaystyle b \cdot x + 0_R$ Power of Identity is Identity $\displaystyle$ $=$ $\displaystyle b \cdot x$ $\displaystyle$ $=$ $\displaystyle \left[\!\left[b\right]\!\right]_m \circ x$ Definition of $\circ$
Thus, the definition of $\circ$ is well-defined.
$\Box$
Let us verify that $\left({R, +, \circ}\right)_{\Z_m}$ is a unitary $\Z_m$-module by verifying the axioms in turn.
### Axiom $(1)$
We need to show that:
$\left[\!\left[{a}\right]\!\right]_m \circ \left({x + y}\right) = {\left[\!\left[{a}\right]\!\right]_m \circ x} + {\left[\!\left[{a}\right]\!\right]_m \circ y}$
$\displaystyle \left[\!\left[a\right]\!\right]_m \circ \left({x + y}\right)$ $=$ $\displaystyle a \cdot \left(x +y\right)$ Definition of $\circ$ $\displaystyle$ $=$ $\displaystyle a \cdot x + a \cdot y$ Power of Product in Abelian Group $\displaystyle$ $=$ $\displaystyle \left[\!\left[a\right]\!\right]_m \circ x + \left[\!\left[a\right]\!\right]_m \circ y$ Definition of $\circ$
$\Box$
### Axiom $(2)$
We need to show that:
$\left( {\left[\!\left[a\right]\!\right]_m +_m \left[\!\left[b\right]\!\right]_m}\right) \circ x = \left[\!\left[a\right]\!\right]_m \circ x + \left[\!\left[b\right]\!\right]_m \circ x$
$\displaystyle \left( {\left[\!\left[a\right]\!\right]_m +_m \left[\!\left[b\right]\!\right]_m}\right) \circ x$ $=$ $\displaystyle \left[\!\left[ a+b \right]\!\right]_m \circ x$ Definition of Modulo Addition $\displaystyle$ $=$ $\displaystyle \left(a+b\right) \cdot x$ Definition of $\circ$ $\displaystyle$ $=$ $\displaystyle a \cdot x + b \cdot x$ Powers of Group Elements: Sum of Indices $\displaystyle$ $=$ $\displaystyle \left[\!\left[a\right]\!\right]_m \circ x + \left[\!\left[b\right]\!\right]_m \circ x$ Definition of $\circ$
$\Box$
### Axiom $(3)$
We need to show that:
$\left({\left[\!\left[a\right]\!\right]_m \times_m \left[\!\left[b\right]\!\right]_m}\right) \circ x = \left[\!\left[a\right]\!\right]_m \circ \left({\left[\!\left[b\right]\!\right]_m \circ x}\right)$
$\displaystyle \left({\left[\!\left[a\right]\!\right]_m \times_m \left[\!\left[b\right]\!\right]_m}\right) \circ x$ $=$ $\displaystyle \left[\!\left[ a \times b \right]\!\right]_m \circ x$ Definition of Modulo Multiplication $\displaystyle$ $=$ $\displaystyle \left(a \times b\right) \cdot x$ Definition of $\circ$ $\displaystyle$ $=$ $\displaystyle a \cdot \left({b \cdot x}\right)$ Powers of Group Elements: Product of Indices $\displaystyle$ $=$ $\displaystyle \left[\!\left[a\right]\!\right]_m \circ \left({\left[\!\left[b\right]\!\right]_m \circ x}\right)$ Definition of $\circ$
$\Box$
### Axiom $(4)$
We need to show that:
$\left[\!\left[1\right]\!\right]_m \circ x = x$
That is, that $1 \cdot x = x$.
This follows from the definition of power of group element.
$\Box$
Having verified all four axioms, we have shown that $\left({R, +, \circ}\right)_{\Z_m}$ is a unitary $\Z_m$-module.
$\blacksquare$
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2020-08-13 15:12:50
|
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http://gmatclub.com/forum/gmat-quantitative-section-7/index-50.html?sk=a&sd=a
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Find all School-related info fast with the new School-Specific MBA Forum
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2015-07-01 21:11:20
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https://www.transtutors.com/questions/ap6-1-now-that-operations-for-outdoor-clinics-and-team-events-are-running-smoothly-s-2637555.htm
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# AP6–1 Now that operations for outdoor clinics and TEAM events are running smoothly, Suzie thinks...
AP6–1 Now that operations for outdoor clinics and TEAM events are running smoothly, Suzie thinks of another area for business expansion. She notices that a few clinic participants wear multiuse (MU) watches. Beyond the normal timekeeping features of most watches, MU watches are able to report temperature, altitude, and barometric pressure. MU watches are waterproof, so moisture from kayaking, rain, fishing, or even diving up to 100 feet won't damage them. Suzie decides to have MU watches available for sale at the start of each clinic. The following transactions relate to purchases and sales of watches during the second half of 2019. All watches are sold for $300 each. Required: 1. (a) Calculate sales revenue, cost of goods sold, and ending inventory as of December 31, 2019, assuming Suzie uses FIFO to account for inventory. (b) Prepare the gross profit section of a partial income statement for transactions related to MU watches. 2. Late in December, the next generation of multiuse (MU II) watches is released. In addition to all of the features of the MU watch, the MU II watches are equipped with a global positioning system (GPS) and have the ability to download and play songs and videos off the Internet. The demand for the original MU watches is greatly reduced. As of December 31, the estimated net realizable value of MU watches is only$100 per watch.
(a) Record any necessary adjustment on December 31, 2019, related to this information.
(b) For what amount would MU inventory be reported in the December 31, 2019, balance sheet?
(c) Prepare an updated gross profit section of a partial income statement accounting for this additional
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. Acid-test ratio. g. Debt to equity ratio. h. Times interest earned ratio. 2 . Calculate the following profitability ratios for 2020. a. Gross profit ratio (on the MU watches ). b . Return on assets. c . Profit margin. d. Asset turnover. e. Return on equity. 3. Briefly
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2018-06-18 02:16:53
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http://math.chapman.edu/~jipsen/structures/doku.php/semigroups_with_zero
|
Semigroups with zero
Abbreviation: Sgrp$_0$
Definition
A semigroup with zero is a structure $\mathbf{S}=\langle S,\cdot,0\rangle$ of type $\langle 2,0\rangle$ such that
$\langle S,\cdot\rangle$ is a semigroups
$0$ is a zero for $\cdot$: $x\cdot 0=0$, $0\cdot x=0$
Morphisms
Let $\mathbf{S}$ and $\mathbf{T}$ be semigroups with zero. A morphism from $\mathbf{S}$ to $\mathbf{T}$ is a function $h:S\rightarrow T$ that is a homomorphism:
$h(x\cdot y)=h(x)\cdot h(y)$, $h(0)=0$
Example 1:
Properties
Classtype variety decidable in PTIME undecidable undecidable no unbounded no no no no no
Finite members
$\begin{array}{lr} f(1)= &1\\ f(2)= &\\ f(3)= &\\ f(4)= &\\ f(5)= &\\ f(6)= &\\ \end{array}$
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2017-03-29 11:07:53
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https://computergraphics.stackexchange.com/questions/8190/how-to-randomly-draw-quaternions-within-a-specific-range-of-euler-angles-for-rot?noredirect=1
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# How to randomly draw Quaternions within a specific range of Euler angles for rotation?
I am pretty new to Quaternions so please bear with me. I want to draw random Quaternion samples so that their Euler angle equivalent would range within [-30, +30] degrees on each axis. Currently, I know how to sample Quaternions from the full range ([-180, +180]) using the code below but I don't know how to modify the code so that I can get samples within the range [-30, +30]. Can anyone help me with that?
I'm not sure if this is helpful to answer this question but here's a piece of information: I eventually want to convert the sampled Quaternion to Euler angles an apply the Euler rotation in some 3D shapes. The order of rotation in the software I'm using to do this is XYZ meaning that it first rotates the 3D shape along the X axis, then Y axis and then Z axis.
import numpy as np
def sample_Quaternion():
r = np.random.uniform(0, 1 - 0.001, 3)
while np.linalg.norm(r) > 1:
r = np.random.uniform(0, 1 - 0.001, 3) # Just to keep the L2 norm within [0, 1.0)
w = [np.sqrt(1 - (r[0]*r[0] + r[1]*r[1] + r[2]*r[2]))]
r = np.concatenate(r, w) # the output of this would represent (x, y, z, w)
return r
• "I eventually want to convert the sampled Quaternion to Euler angles an apply the Euler rotation in some 3D shapes." Please stop wanting to do that. Euler angles are bad, and you've taken a big step by moving away from them. Keep your orientations as quaternions; don't convert to/from Euler angles. Oct 28 '18 at 1:48
• @NicolBolas I really want to avoid using Euler angles but unfortunately I have to use Euler angles for a part of my current project. Could you also take a look at my new, relevant question here and see if you can provide an answer for this one?
– Amir
Oct 28 '18 at 16:57
• @Amir Why do not you just use matrices to rotate? Nov 2 '18 at 2:51
The simplest way to do this would be to generate the Euler angles randomly, then use them to create the quaternion. Most existing quaternion implementations in game engines and 3D math libraries will have a FromEuler ( Vector3 ) function.
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2022-01-22 15:50:52
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https://tex.stackexchange.com/questions/475357/why-does-adding-a-bar-in-a-differential-equation-cause-all-these-errors
|
Why does adding a \Bar in a differential equation cause all these errors?
The line in question:
B &= \diffp* { f(x, \Bar{\mu}) } { \Bar{\mu} } { x_{eq}, \Bar{\mu}_{eq} }
The second \Bar{\mu} (bolded for clarity) seems to throw everything awry. Omitting the \Bar (just having '{ \mu }') works just fine, everything compiles with no errors and displays as expected. Including the \Bar gives me 44 errors, the first few of which would suggest I forgot a '}', but I've checked and triple checked, and I can find no missing or mismatched brackets.
I'm guessing there's some quirk of \diffp* that's causing this, but it's what I've used everywhere else in the document, and I'd rather not have to go back and change them all to \frac for consistency. So how do I get a bar on the bottom with \diffp*, and why is my way incorrect?
EDIT: \diffp* comes from the package esdiff.
A full example doc:
\documentclass[12pt]{extarticle}
\usepackage[utf8]{inputenc}
\usepackage{cite}
\usepackage{amsmath}
\usepackage{esdiff}
\usepackage{ amssymb }
\begin{document}
\begin{align*}
A &= \diffp*{f(x,\bar{\mu})}{x}{x_{eq},\Bar{\mu}_{eq}}\\
B &= \diffp* {f(x, \Bar{\mu})} {\Bar{\mu}} {x_{eq},\Bar{\mu}_{eq}}
\end{align*}
\end{document}
• What is \Bar? Where did you get it from? What package? Did you mean \bar? – Steven B. Segletes Feb 17 '19 at 17:36
• Definitely meant \Bar. I'm pretty sure it's from amsmath, could be amssymb. I should probably also mention that this line is in an align* block with a similar line above it. – Nick J Feb 17 '19 at 17:44
• After double checking, \bar and \Bar seem to do the same thing. Both add a bar no top of the symbol, both throw the same errors in that line. – Nick J Feb 17 '19 at 17:47
• Why don't you give us a small, but complete document that displays the issue? – Steven B. Segletes Feb 17 '19 at 17:51
• \documentclass[12pt]{extarticle} \usepackage[utf8]{inputenc} \usepackage{cite} \usepackage{amsmath} \usepackage{esdiff} \usepackage{ amssymb } \begin{document} \begin{align*} A &= \diffp*{f(x,\bar{\mu})}{x}{x_{eq},\Bar{\mu}_{eq}}\\ B &= \diffp* {f(x, \Bar{\mu})} {\Bar{\mu}} {x_{eq},\Bar{\mu}_{eq}} \end{align*} \end{document} – Nick J Feb 17 '19 at 17:54
You seem to have missed out the {} of one argument, but some failing in the \diffp macro requires it to be doubled.
\documentclass[12pt]{extarticle}
\usepackage[utf8]{inputenc}
\usepackage{cite}
\usepackage{amsmath}
\usepackage{esdiff}
\usepackage{ amssymb }
\begin{document}
\begin{align*}
A &= \diffp*{f(x,\bar{\mu})}{x}{x_{\mathrm{eq}},\Bar{\mu}_{\mathrm{eq}}}\\
B &= \diffp* {f(x, \Bar{\mu})} {{\Bar{\mu}}} {x_{\mathrm{eq}},\Bar{\mu}_{\mathrm{eq}}}
\end{align*}
\end{document}
The problem comes from the second equation B &= \diffp* {f(x, \Bar{\mu})} {\Bar{\mu}} {x_{eq},\Bar{\mu}_{eq}}. The second argument \Bar{\mu} should be in parenthesis, ie, {\Bar{\mu}}. So, second equation should be B &= \diffp* {f(x, \Bar{\mu})} {{\Bar{\mu}}} {x_{eq},\Bar{\mu}_{eq}}
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2020-06-05 18:29:47
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https://www.physicsforums.com/threads/ratio-of-atoms-in-different-laser-levels.841208/
|
# Ratio of atoms in different laser levels
1. Nov 3, 2015
### Jon.G
1. The problem statement, all variables and given/known data
For large intensities, what is the maximum ratio of atoms in N1 compared with N3?
I suppose this is just a bit of maths I'm struggling with, seems simple I just can't get it for whatever reason D:
Working through previous question, I got to:
N3= (Ip/Is) / ((Ip/Is) +1 ) N1
(Ip is the 'pump' intensity, Is is the saturation intensity)
2. Relevant equations
3. The attempt at a solution
My first thought was with the large intensities, they would be >> 1, so I could just ignore that and get N3 = N1. But then because it's the same fraction in the numerator and the denominator, and the +1 comes after the Ip/Is, I thought this to be wrong.
My only other thought is that if the fraction reaches the point where it becomes 1, then N3 = 0.5 N1.
But I'm not sure where my reasoning of this would come from. I don't think it would be ∞/∞ as, if I am correct, that is not defined, not 1.
Any hints?
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2017-11-23 00:22:42
|
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|
https://www.albert.io/ie/trigonometry/area-triangle-law-cosines-three-known-sides
|
?
Free Version
Difficult
Area, Triangle, Law Cosines, Three Known Sides
TRIG-WV7OAE
If you are given a triangular piece of wood with the following dimensions, $11\;cm,\;18\;cm\;$ and $\;20\;cm$, what is the area of that piece of wood to the nearest square cm?
A
$98.4\; sq.\; cm$
B
$150.8\; sq.\; cm$
C
$117.4\; sq.\; cm$
D
$178.3\; sq.\; cm$
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2016-12-08 08:01:39
|
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https://www.physicsforums.com/threads/book-on-functional-calculus.983967/
|
# Book on Functional Calculus
• Calculus
## Main Question or Discussion Point
Hi, I'm looking for a book that explains more deeply (and a little bit more formal) the functional calculus than the typical introductions that I find in QFT books (like Peskin or Hatfield). Is there any good book for physicists to learn the mathematics behind functional calculus?
Thanks
## Answers and Replies
Related Science and Math Textbooks News on Phys.org
MathematicalPhysicist
Gold Member
How about Dewitt's Functional Integration.
I must confess that I haven't read this book yet, but looks like a good place to start.
Ok, thanks, although it's not quite what I was looking for, I'm more interested in a book that starts without assuming any functional knowledge and starts constructing what a functional is, it's derivatives, functional equations etc...
MathematicalPhysicist
Gold Member
So perhaps you need a book in Calculus of Variations?
I prefer rigorous maths (when possible), so my recommendations will be as such.
Yes, if the topic is discussed in those books. I also like mathematical rigour, although I'm not a mathematician and I usually look for a trade-off between rigour and not have to read 20 books before to understand the first statement :D.
MathematicalPhysicist
Gold Member
Yes, if the topic is discussed in those books. I also like mathematical rigour, although I'm not a mathematician and I usually look for a trade-off between rigour and not have to read 20 books before to understand the first statement :D.
Well if the topic is so advanced then perhaps those 20 books are in order...
Anyway, a good place to start is An Introduction to Variational Calculus by Bernard Darcogona (google will correct my spelling). I read through chapters 1-4 didn't finish it though.
Mind you, that you need to know before that Hilbert spaces and intro to Functional Analysis.
You know all those spaces of $L^p$ etc.
vanhees71
Science Advisor
Gold Member
2019 Award
The classic still is Hilbert and Courant vol. I (not the modenized single-volume rewriting of it, which destroys the charm of the original at least somewhat).
MathematicalPhysicist
Gold Member
The classic still is Hilbert and Courant vol. I (not the modenized single-volume rewriting of it, which destroys the charm of the original at least somewhat).
yes, also that book I never finished reading.
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2020-02-26 23:50:47
|
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http://jasontdean.com/python/wine_classification.html
|
## Wine Classification¶
Jason Dean
July 6th, 2017
jtdean@gmail.com
https://github.com/jtdean123
Here we build a simple decision tree to predict whether a wine is white or red based on the following attributes:
● Fixed acidity
● Free sulphur dioxide
● Volatile acidity
● Total sulphur dioxide
● Citric acid
● Residual sugar
● pH
In [117]:
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from sklearn.tree import DecisionTreeClassifier as DTC
from sklearn.metrics import confusion_matrix
from sklearn.model_selection import train_test_split
from sklearn.model_selection import RandomizedSearchCV
import itertools
import time
from sklearn.tree import export_graphviz
import subprocess
from IPython.display import Image
### Exploratory Data Analysis¶
In [2]:
# read the data
Out[2]:
fixed_acidity volatile_acidity citric_acid residual_sugar free_sulfur_dioxide total_sulfur_dioxide ph class
0 7.4 0.70 0.00 1.9 11.0 34.0 3.51 1
1 7.8 0.88 0.00 2.6 25.0 67.0 3.20 1
2 7.8 0.76 0.04 2.3 15.0 54.0 3.26 1
3 11.2 0.28 0.56 1.9 17.0 60.0 3.16 1
4 7.4 0.70 0.00 1.9 11.0 34.0 3.51 1
In [3]:
# quick sanity check to make sure we have numerical data and not str
for i in range(0, wine.shape[1]):
for j in range(0, wine.shape[0]):
if type(wine.iloc[j,i]) is str:
print 'type str at : ', j, i
print 'congrats, no strings'
congrats, no strings
In [4]:
type(wine.iloc[3,2])
Out[4]:
numpy.float64
We have 7 features avaialable to us, and to start we will use all of them. Next we determine the number of observations in the data set.
In [5]:
print 'Number of observations: ', wine.shape[0]
Number of observations: 6497
There are 6497 wine observations and each one is either white or red. Red wine is indiciated by a class value of '1' and white wine by '0'.
In [6]:
print 'Number of red wine observations: ', wine[wine['class'] == 1].shape[0]
print 'Number of white wine observations: ', wine[wine['class'] == 0].shape[0]
Number of red wine observations: 1599
Number of white wine observations: 4898
It appears this is a moderately unbalanced data set - there are exactly 3 times as many white wine observations are there are red. Next, like the good data scientists we are, we check for missing data.
In [7]:
wine.isnull().values.any()
Out[7]:
False
No missing data - it appears that wine was not being consumed during data entry. We will also check for duplicated data entries.
In [8]:
wine.duplicated().values.any()
Out[8]:
True
Some of the observations, or rows, in the data frame are duplicated. Since there is not a unique wine identifier this may not be a problem - it is possible for two different wines to have identical values for each of the features. However, if a red and white wine have identical values for all of the features then this would introduce problems during model buidling. Now that have checked for missing data we can start exploring the data. We start with summary statistics of each feature.
In [9]:
# evaluate mean grouped by class
wine.groupby(['class']).mean()
Out[9]:
fixed_acidity volatile_acidity citric_acid residual_sugar free_sulfur_dioxide total_sulfur_dioxide ph
class
0 6.854788 0.278241 0.334192 6.391415 35.308085 138.360657 3.188267
1 8.319637 0.527821 0.270976 2.538806 15.874922 46.467792 3.311113
As shown above, red wine has lower fixed and volatile acidity, higher citric acid, residual sugar, free and total SD, and, consistent with these measurments, lower mean pH. Taken together, this data indicates that red wine is, on average, more acidic and contains more sugar than white wine. The mean as a summary statistic can be misleading if a feature is skewed, so we next evaluate the median.
In [10]:
wine.groupby(['class']).median()
Out[10]:
fixed_acidity volatile_acidity citric_acid residual_sugar free_sulfur_dioxide total_sulfur_dioxide ph
class
0 6.8 0.26 0.32 5.2 34.0 134.0 3.18
1 7.9 0.52 0.26 2.2 14.0 38.0 3.31
The median values for these features are similar to the mean, indicating that there is not significant skew (maybe...) in the data and that the conclusions we stated above with regards to acidity seem valid. Since is seems that pH and sugar content are key differentiators between red and white wine we will visualize them with histograms. We start with fixed acidity, a measure of nonvolatile acids.
In [14]:
# pH
red_pH = list(wine[wine['class'] == 1]['fixed_acidity'])
white_pH = list(wine[wine['class'] == 0]['fixed_acidity'])
font = {'weight' : 'bold',
'size' : 18}
plt.rc('font', **font)
weights_r = np.ones_like(red_pH)/len(red_pH)
weights_w = np.ones_like(white_pH)/len(white_pH)
bins = np.arange(0,14,0.2)
plt.figure(figsize=(10, 6))
plt.hist(red_pH, bins, alpha=0.5, label='red', color='red', edgecolor = "black", weights=weights_r)
plt.hist(white_pH, bins, alpha=0.5, label='white', color='lightblue', edgecolor = "black", weights=weights_w)
plt.legend(loc='upper right')
plt.xlabel('fixed_acidity')
plt.ylabel('Fraction')
plt.show()
plt.rcdefaults()
In [13]:
# sugar
red_sugar = list(wine[wine['class'] == 1]['residual_sugar'])
white_sugar = list(wine[wine['class'] == 0]['residual_sugar'])
font = {'weight' : 'bold',
'size' : 18}
plt.rc('font', **font)
weights_r = np.ones_like(red_pH)/len(red_pH)
weights_w = np.ones_like(white_pH)/len(white_pH)
bins = np.arange(0,wine['residual_sugar'].max(),0.5)
plt.figure(figsize=(10, 6))
plt.hist(red_sugar, bins, alpha=0.5, label='red', color='red', edgecolor = "black", weights=weights_r)
plt.hist(white_sugar, bins, alpha=0.5, label='white', color='lightblue', edgecolor = "black", weights=weights_w)
plt.legend(loc='upper right')
plt.xlabel('residual_sugar')
plt.ylabel('Fraction')
plt.show()
plt.rcdefaults()
Interestingly, although we found that red wine has a highher mean and median residual sugar content than white wine we see that the residual sugar distribution for white wine shows a right skew, indicating that some types of white wine contain much more sugar than others. Riesling! Finally, we examine the relationship between residual sugar and pH.
In [15]:
# sugar vs. pH
font = {'weight' : 'bold',
'size' : 18}
plt.rc('font', **font)
plt.figure(figsize=(10, 6))
ax = wine[wine['class']==0].plot(kind='scatter', x='residual_sugar', y='fixed_acidity', color='red', label='red');
wine[wine['class']==1].plot(kind='scatter', x='residual_sugar', y='fixed_acidity', color='lightblue', label='white', ax=ax);
plt.show()
<matplotlib.figure.Figure at 0x113bcc150>
These two features are not great separators of the two classes, but fortunately for us we have 5 others to include in the model. Sadly I am not able to visualize data in 7 dimensions so we will have to let a decision tree classify for us..
### Decision Tree Model Building¶
The goal of this project is to build a decision tree to classify wine as white or red based on the features that we have available to us. Feature engineering, the process of eliminating, combining, or transforming features prior to model construction seems to be the hardest and most important part of a machine learning task. How do we identify what features to include in this model? The easiest place to start is to include all of them, so that is what we will do.
First we will build a decision tree using default parameters and all of the features. Before we begin, though, we will split the data into a test and training set.
In [16]:
# split the data into test and training sets
wine_train, wine_test, class_train, class_test = train_test_split((wine.drop(['class'], axis=1)),
wine['class'],
test_size=0.3,
random_state=321)
In [17]:
# make a few functions that will be useful for model building
def accuracy(actual, predictions):
'''
calculate the accuracy as: # correct / # total
'''
correct = 0
for i,j in zip(list(actual), predictions):
if i == j:
correct += 1.
return correct/len(actual)
def confusion(actual, predictions):
'''
generate a confusion matrix
'''
confusion_mat = pd.DataFrame(confusion_matrix(actual, predictions))
confusion_mat.columns = ['Predicted White', 'Predicted Red']
confusion_mat.index = ['Actual White', 'Actual Red']
return confusion_mat
def area_under_curve(actual, predictions):
'''
calculate the auc under a roc using simpson's method
'''
tp=0; fp=0; tn=0;
for i, j in zip(actual, predictions):
if j == 1 and i == 1:
tp += 1.
if j == 1 and i == 0:
fp += 1.
if j == 0 and i == 0:
tn += 1.
# calcuate true positive and false positive rates
tpr = tp/(tp + fp)
fpr = fp/(fp + tn)
# calculate area under the curve
area = 1/2 * fpr* tpr + (1-fpr) * tpr + 1/2 * (1-tpr)*(1-fpr)
return area
In [18]:
# build a decision tree with all of the features
tree_all = DTC()
tree_all_DCT = tree_all.fit(wine_train, class_train)
preds_all = tree_all_DCT.predict(wine_test)
print 'Accuracy: ', accuracy(class_test, preds_all)
confusion(class_test, preds_all)
Accuracy: 0.961538461538
Out[18]:
Predicted White Predicted Red
Actual White 1437 41
Actual Red 34 438
Not bad! These practice data sets are good for my ego. As shown above, without doing any cross validation or hyperparameter tuning we achieved 96.1% accuracy on the test set. Since this is an unbalanced data set the accuracy is not that informative, however, so next we calculate the AUC. The AUC is defined as the area under a ROC curve. The ROC curve is a plot of true positive rate vs false positive rate, and for a non-probabilistic model like a decision tree the AUC is defined by a single point.
The true positive rate is defined as: true positive / (true positive + false positive)
The false positive rate is defined as: false positive / (false positive + true negative)
Therefore, to calcualte the AUC we determined the area under a curve bounded by [0,0], [tpr, fpr], and [1,1] and we will do this using basic geometry by adding the area of two triangles and a rectange.
In [19]:
print 'AUC for model with all features included: ', area_under_curve(class_test, preds_all)
AUC for model with all features included: 0.889039242219
As shown above, we achieved an AUC of 0.89 by using default parameters for the decision tree and using all of the parameters. In order to better serve the wine community we will try to improve our model by doing some feature engineering. Since we only have 7 features it is feasible to try all combinations of them... and that is what we will do! Plus cross validation (RandomizedSearchCV). We first make create functions to perform CV, generate a model, and return an AUC.
In [57]:
def cross_validation_model(x_train, x_class, y_test, y_class):
'''
generate a model via 10x CV and return an AUC
'''
# set of parameters to test
param_grid = {"criterion": ["gini", "entropy"],
"min_samples_split": [2, 5, 10, 20, 50],
"max_depth": [None, 2 , 4, 5, 10, 20],
"min_samples_leaf": [1, 2, 3, 5, 10, 20, 50],
"max_leaf_nodes": [None, 5, 10, 20 , 50],
}
# gridsearch CV 10 fold
dtc = DTC()
clf = RandomizedSearchCV(dtc, param_grid, cv=10)
clf.fit(x_train, x_class)
# create a model with the best parameters and predict
preds = clf.predict(y_test)
return [area_under_curve(y_class, preds), accuracy(y_class, preds)]
Before we try feature engineering we will test if just cross validation is enough to improve our AUC.
In [59]:
# wine_train, wine_test, class_train, class_test
all_features_cv = cross_validation_model(wine_train, class_train, wine_test, class_test)
print 'AUC, all features, 10 fold CV: ', all_features_cv[0]
print 'Accuracy, all features, 10 fold CV: ', all_features_cv[1]
AUC, all features, 10 fold CV: 0.919368186913
Accuracy, all features, 10 fold CV: 0.967179487179
We improved the accuracy and AUC of the model by performing 10 fold CV to hyperparameter tune the model. We now evaluate the accuracy and AUC of all possible feature combinations.
In [60]:
# evaluate all possible combinations of the 7 features
combinations = []
features = [0, 1, 2, 3, 4, 5, 6]
for i in features:
combos = list(itertools.combinations(features, i))
for j in combos:
combinations.append(j)
print 'total # of combinations: ', len(combinations)
total # of combinations: 127
In [61]:
# generate all possible models and evaluate performance
start_time = time.time()
combos_performance = []
for i in combinations:
if len(i) == 0: continue
i = list(i)
combos_train = wine_train.iloc[:, i]
combos_test = wine_test.iloc[:, i]
combos_cv = cross_validation_model(combos_train, class_train, combos_test, class_test)
combos_performance.append(combos_cv)
print("--- %s seconds ---" % (time.time() - start_time))
--- 101.671665907 seconds ---
In [73]:
# find the model that generate the higest accuracy and plot
auc_combos = [i[0] for i in combos_performance]
acc_combos = [i[1] for i in combos_performance]
plt.scatter(auc_combos, acc_combos, edgecolors='black')
plt.xlabel('Accuracy')
plt.ylabel('AUC')
plt.grid()
plt.show()
Some of the models are relatively bad, but we see what appears to be a linear trend between AUC and accuracy. Now we can find the model that generated the best accuracy.
In [126]:
# find the best model
max_acc = 0; index=0;
for i, j in enumerate(acc_combos):
if j > max_acc:
max_acc = j
index = i
# determine what features were included in this model
print 'Features included in best model: ', combinations[index]
print 'Best accuaracy: ', max_acc
Features included in best model: (0, 1, 3, 4)
Best accuaracy: 0.972820512821
#### We find that the best model includes four features: fixed_acidity, volatile_acidy, residual_sugar, and free_sulfur_dioxide.¶
Finally, we can visualize the best decision tree.
In [112]:
# set of parameters to test
#wine_train, wine_test, class_train, class_test
param_grid = {"criterion": ["gini", "entropy"],
"min_samples_split": [2, 5, 10, 20, 50],
"max_depth": [None, 2 , 4, 5, 10, 20],
"min_samples_leaf": [1, 2, 3, 5, 10, 20, 50],
"max_leaf_nodes": [None, 5, 10, 20 , 50],
}
# gridsearch CV 10 fold, then generate a model w/ the best parameters
dtc = DTC()
clf = RandomizedSearchCV(dtc, param_grid, cv=10)
clf.fit(wine_train.iloc[:,list(combinations[index])], class_train)
dtc = DTC(criterion=clf.best_estimator_.criterion,
min_samples_split = clf.best_estimator_.min_samples_split,
max_depth = clf.best_estimator_.max_depth,
min_samples_leaf = clf.best_estimator_.min_samples_leaf,
max_leaf_nodes = clf.best_estimator_.max_leaf_nodes)
dtc.fit(wine_train.iloc[:,list(combinations[index])], class_train)
# display the tree!
output_dot = 'decision_tree.dot'
feature_names = ['fixed_acidity', 'volatile_acidy', 'residual_sugar', 'free_sulfur_dioxide']
with open(output_dot, 'w') as f:
f = export_graphviz(dtc, out_file=f, feature_names=feature_names)
# paste the output into-
In [123]:
%%bash
dot -Tpng decision_tree.dot -o decision_tree.png
In [125]:
Image("decision_tree.png")
Out[125]:
|
2021-10-19 16:07:24
|
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|
http://math.stackexchange.com/questions/285313/prove-lim-x-rightarrow-0-frac-sinxx-1-with-the-epsilon-delta-def
|
# Prove $\lim_{x \rightarrow 0} \frac {\sin(x)}{x} = 1$ with the epsilon-delta definition of limit.
It is well known that
$$\lim_{x \rightarrow 0} \frac {\sin(x)}{x} = 1$$
I know several proofs of this: the geometric proof shows that $\cos(\theta)\leq\frac {\sin(\theta)}{\theta}\leq1$ and using the Squeeze Theorem I conclude that $\lim_{x \rightarrow 0} \frac {\sin(x)}{x} = 1$, other proof uses the Maclaurin series of $\sin(x)$. My question is: is there a demonstration of this limit using the epsilon-delta definition of limit?
-
A proof with Maclaurin series sounds somehow debatable or at least strange - you need derivatives to write this series and the question itself is about the derivative of $\sin$ at $0$. – savick01 Jan 23 '13 at 20:31
what definition of sin – user58512 Jan 23 '13 at 20:32
I think the infinite sum is more rigourous definition of sine function than geometric hand waving. It is not debatable. – user123454321 Jan 23 '13 at 20:32
To expand what @user58512 says, if you use a geometric definition of $sin x$, then you should expect a proof with a geometrical basis. One could modify the proof using the Squeeze Theorem to conform to the epsilon-delta definition. So again, what definition of sine? – Michael E2 Jan 23 '13 at 20:39
@MichaelE2 the infinite sum. – dwarandae Jan 23 '13 at 20:41
Here is a more direct answer for this: Since $\cos\theta<\frac{\sin\theta}{\theta}<1$, one can get $$\big|\frac{\sin\theta}{\theta}-1\big|<1-\cos\theta.$$ But $1-\cos\theta=2\sin^2\frac{\theta}{2}\le\frac{\theta^2}{2}$ and hence $$\big|\frac{\sin\theta}{\theta}-1\big|\le\frac{\theta^2}{2}.$$ Now it is easy to use $\varepsilon-\delta$ definition to get the answer.
-
For every $x \ne 0$ we have $$\Big|1-\frac{\sin x}{x}\Big|=\Big|1-\sum_{k=0}^\infty\frac{(-1)^kx^{2k}}{(2k+1)!}\Big|\le \sum_{k=1}^\infty\frac{|x|^{2k}}{(2k+1)!}\le\frac13\sum_{k=1}^\infty\frac{|x|^{2k}}{(2k)!}=\frac{\cosh|x|-1}{3}.$$ Given $\varepsilon>0$, let $\delta=\cosh^{-1}(1+3\varepsilon)$. Then $$0<|x|\le\delta \Longrightarrow \Big|1-\frac{\sin x}{x}\Big|\le\frac{\cosh|x|-1}{3}\le \varepsilon.$$ Another approach is to notice that $$x-\sin x\le \frac{x^2}{2} \quad \forall\ x \in [0,\pi].$$ Since $\sin$ is odd we have $$-x+\sin x\le \frac{x^2}{2} \quad \forall\ x \in [-\pi,0].$$ Hence $$|x-\sin x|\le \frac{x^2}{2} \quad \forall\ x \in [-\pi,\pi].$$ Given $\varepsilon>0$, we have $$0<|x|\le 2\varepsilon \Longrightarrow \Big|1-\frac{\sin x}{x}\Big|\le\frac{|x|}{2} \le \varepsilon.$$
-
It seems to me that there is a big problem with using the Taylor series. Notice that $$\frac{d}{dx} \sin x := \lim_{h \to 0} \frac{\sin(x+h)-\sin x}{h} \equiv \lim_{h \to 0} \left[ \left(\frac{\cos h -1}{h}\right) \sin x+ \left(\frac{\sin h}{h}\right) \cos x \right].$$ By using the Taylor series, you are using the fact that the derivative of $\sin x$ is $\cos x$, and so are tacitly assuming that $(\sin h)/h \to 1$ as $h \to 0$. – Fly by Night Jan 23 '13 at 21:06
@Fly by Night One way to define sine is to define it by its Taylor series. (This makes the problem easier though). – Amr Jan 23 '13 at 21:33
@Amr Indeed, in that case the proof of the limit is trivial (it becomes an application of the laws of indices). Moreover, the need to know the limit of $\frac{1}{x}\sin x$ also disappears because we can calculate $\frac{d}{dx}\sin x$ by differentiating monomials. The "definition" of $\sin x$ as a series comes from Taylor series and comes from assuming that $\frac{1}{x}\sin x \to 1$ as $x \to 0$ and then pretending to forget where we got the series from. A little dishonest in my opinion. – Fly by Night Jan 23 '13 at 21:37
@Fly by Night I agree. I will try to put a solution (Its long though) – Amr Jan 23 '13 at 21:51
@FlybyNight: there is more than one way to define $\sin(x)$. One is by the geometric ratio. Another is by its power series (which is being misnamed Taylor Series here). If defined as a power series, $\frac{\sin(x)}{x}$ is just a power series whose constant term is $1$ and the result is trivial. We don't need to worry about evaluating $$\lim_{x\to0}\frac{\sin(x+h)-\sin(x)}{h}$$ using any trigonometric formulas. – robjohn Jan 24 '13 at 0:42
I believe you are looking for a a proof without differentiation but only metric spaces.
Define $e^z = \sum_{n = 0}^\infty \frac{z^n}{n!}$, define $sin(z) = \frac{1}{2i}(e^{iz}-e^{-iz})$
$\frac{\sin(z)}{z} = 1 +\sum_{k =1 }^\infty (-1)^k \frac{z^{2k}}{(2k+1)!}$
Now we prove the sum of the latter term goes to 0:
$0\le|\sum_{k =1 }^\infty (-1)^k \frac{z^{2k}}{(2k+1)!}|\le \sum_{k =1 }^\infty |\frac{z^{2k}}{(2k+1)!}|\le \sum_{k =1 }^\infty |\frac{z^{2k}}{6^{2k}}|=\sum_{k =1 }^\infty |\frac{z}{6}|^{2k}$
As $z\rightarrow 0$, we can pick $\displaystyle N\in\mathbb{N}.\;\forall n\ge N.\;|z_n|<6\;\Longrightarrow |\frac{z_n}{6}|<1$
$\lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}} \sum_{k =1 }^\infty |\frac{z}{6}|^{2k} = \lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}}(\frac{1}{1-|\frac{z}{6}|}-1) = 1-1 =0$
Here we are comparing it with geometric sum. When its absolute value is sandwiched between $0$, the term has to go to $0$.
$\Longrightarrow$ $0\le\lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}}$$|\sum_{k =1 }^\infty (-1)^k \frac{z^{2k}}{(2k+1)!}| \le \lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}} \sum_{k =1 }^\infty |\frac{z}{6}|^{2k}=0 \Longrightarrow \lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}}\sum_{k =1 }^\infty (-1)^k \frac{z^{2k}}{(2k+1)!} = 0 \Longrightarrow \lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}}\frac{\sin(z)}{z} = 1 +\lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}}\sum_{k =1 }^\infty (-1)^k \frac{z^{2k}}{(2k+1)!} = 1 - The definition of e^z needs some justification. Also, doesn't the proof that e^{iz} = \cos z + i \sin z rely on Taylor series and therefore involve differentiation? – Fly by Night Jan 23 '13 at 21:09 It depends on how you build your theory. e^z needs to be proven uniform convergent in \mathbb{C} to begin with, then if you define cos and sin with exp, then you don't need to prove Euler's formula, because you used it in definition. What need proven is existence of \pi, periodicity and such so that this cos and sin defined matches the geometric interpretation of what we learnt in high school, which is troublesome. – mez Jan 23 '13 at 21:22 But then surely, if we define \sin z = \frac{1}{2i}(e^{iz}-e^{-iz}) then the fact that (\sin z)/z \to 1 as z \to 0 is trivial, and moreover, the need to know the limit of (\sin z)/z disappears because we can calculate \frac{d}{dz} \sin z from its series definition. – Fly by Night Jan 23 '13 at 21:32 @FlybyNight: actually, you can prove that \cos(x)+i\sin(x)=e^{ix}=\lim\limits_{n\to\infty}\left(1+\frac{ix}{n}\right)^n as shown here. – robjohn Jan 23 '13 at 22:14 @robjohn In equation [7] you use the fact that (\tan x)/x \to 1 as x \to 0. This is equivalent to proving that (\sin x)/x \to 1 as x \to 0. We're back to the same old problem. – Fly by Night Jan 24 '13 at 16:43 Define \sin(x) := x - x^{3}/3! + x^{5}/5! - \cdots and show that this is an analytic function and see that we can take derivative term by term so that \sin'(x) = 1 + x \cdot f(x) for some continuous f. We have the required limit = \sin'(0) = 1. This does not involve any geometric argument and you can trace all the process until you meet \epsilon and \delta. ALSO: See Walter Rudin's Principles of Mathematical Analysis Theorem 8.1. - The definition of \sin x needs some justification. – Fly by Night Jan 23 '13 at 21:08 @FlybyNight And since it is easy to see that the sum is convergent for all real x, I don't see the necessity for "justification of definition," because it is well-defined. Are you talking about relating more intuitive and geometric definition to this one? Even OP commented that he/she is using summation definition. What is there to justify? Only thing that you should justify is whether you can take term by term derivative, which you can find in the reference. – user123454321 Jan 23 '13 at 21:22 @FlybyNight Note that I did not use cosine. In the proof, we have f(x) = -(x^{2}/3! - x^{4}/5 + \cdots). There is no need for quotation marks. – user123454321 Jan 23 '13 at 21:26 A definition does not need justification. We can define f(x)=\sum_{i=1}^{\infty}(-1)^i x^{(2i+1)}/(i+1)! at any point. The justification comes when we want to say that f(x)=\sin (x) when someone else has already picked a definition for \sin(x). But if there is no preexisting definition then there can be no contention as to how I choose to define \sin(x). I think there are problems with definition the function \sin(x) from geometry, because you're already leaning on equivalence of angles mod 2 \pi to able to extend the domain to all real numbers. – R R Mar 15 '14 at 22:53 Then there is the question of extending the domain to all complex numbers from the geometric definition which I have never seen done. \sin(x) exists as an analytical animal first and foremost when we treat it as a function. – R R Mar 15 '14 at 22:54 Taking as the definition per the OP's comment,$$\sin x = \sum_{n=0}^\infty {(-1)^n x^{2n+1} \over (2n+1)!}$$we have that$${\sin x \over x} = \sum_{n=0}^\infty {(-1)^n x^{2n} \over (2n+1)!} =1 - {x^2 \over 6} + {x^4 \over 120} - \cdots \,,$$which is an alternating series for all x and with decreasing terms for, say, |x| < 1. Ok, so let \epsilon>0 be arbitrary. Let \delta = {\rm min}\{\sqrt{6\epsilon},1\}. Assume |x| < \delta. Then$$\left\vert 1 - {\sin x \over x}\right\vert \le {x^2 \over 6}$$since the series is alternating with decreasing terms, and therefore$$\left\vert 1 - {\sin x \over x}\right\vert \le {x^2 \over 6}< {\big(\sqrt{6\epsilon}\big)^2 \over 6}=\epsilon$$Thus the limit is$1\$.
-
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2015-07-07 20:00:38
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https://quantumchemistryniser.wordpress.com/2013/05/25/potential-energy-profile-using-pgfplots/
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# Potential energy profile using PGFplots
Whether you are a master’s student or graduate student you may have to write a research paper one day. To get it published in a good journal, presentation of data has become very important these days until unless you are well known in your field. In this blog i will be doing a small tutorial on how to make a good looking potential energy profile for a well known Diels- Alder reaction. Diels Alder reaction is happening between 1,3-butadiene and acrylo nitrile.
There are different ways of making this energy level diagram using softwares such as : GNUPLOT, Grace, Origin ets. But the problem with all these softwares is the user friendliness. My experience is that you can make any plot using GNUPLOT but you need to be a master in that. Tikz package from LATEX came as boon for me. It is very easy if you are familiar with LATEX and gives figures in PDF format with very high resolution like the one shown below. If you are familiar with latex you will be easily able to work with PGFplots. here is the figure and latex code is given below.
B3LYP/6-31g* level of theory is used for studying the Diels Alder reaction and the energies of stationary points are given in the figure. Latex code for generating the plots is given below. First generate the figures of reactants, product and transition state using Molden in EPS format. This will keep the resolution intact. Place these figures at the nodes corresponding to the reactant, TS and product.
% Diels Alder Reaction - Potential energy Profile
% Author: Leela Sriram Dodda
\documentclass{article}
\usepackage{pgfplots}
\usepackage{tikz}
\usepackage{verbatim}
\usepackage[active,tightpage]{preview}
\PreviewEnvironment{center}
\setlength\PreviewBorder{10pt}%
\begin{document}
\begin{center}
\begin{tikzpicture}[scale=1.0]
\begin{axis}[clip=false,ylabel=Potential energy (kcal/mol),xlabel=Rxn coordinate,xtick=\empty,
legend pos=outer north east,xmin=1,xmax=8,ymax=60,
title=Energy profile for Diels-Alder Reaction,
]
%\node at (2.5,3) {\includegraphics[scale=0.1]{rea1.pdf}};
\node at (axis cs:2.1,-10) {\includegraphics[scale=0.06,angle=90]{r1.pdf}};
\node at (axis cs:3.2,-10) {\includegraphics[scale=0.06,angle=90]{r2.pdf}};
\node[font=\tiny] at (axis cs:2.55,-10) {\bf{+}};
\addplot[color=red,draw=black,line width=0.8pt,densely dotted] coordinates { (3,0)(4,38.4) };
\addplot[color=red,draw=black,line width=0.8pt,densely dotted] coordinates { (5,38.4)(6,-45.9) };
\addplot[color=red,draw=blue,line width=1pt] coordinates {(2,0) (3,0) };
\addplot[color=red,draw=blue,line width=1pt] coordinates {(4,38.4) (5,38.4) };
\node at (axis cs:4.3,50) {\includegraphics[scale=0.08,angle=90]{t1.pdf}};
\node at (axis cs:6.9,-35) {\includegraphics[scale=0.08,angle=90]{p1.pdf}};
\addplot[color=red,draw=blue,line width=1pt] coordinates {(6,-45.9) (7,-45.9) };
\end{axis}
\end{tikzpicture}
\end{center}
\end{document}
This worked for me like a magic hope this will work for you too 🙂
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2017-03-27 10:31:19
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https://www.bbc.co.uk/bitesize/guides/z26nb9q/revision/1
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# Finding volume by counting cubes
The volume of a shape measures the three dimensional (3D) amount of space it takes up. Volume is measured in cubes.
A cubic centimetre is the volume within a cube that has sides of length $${1~cm}$$, as shown above. It has a volume of $${1~cm}^3$$ ($$1~cm$$ cubed).
This cuboid contains $$12$$ cubes. Each cube has a volume of $${1~cm}^3$$. So the volume of this cuboid is $${12~cm}^3$$.
Question
Find the volume of the following cuboids:
a)
b)
a) $${8~cm}^3$$
b) There are two layers of nine cubes, so the volume is $${18~cm}^3$$.
Question
Find the volume of the following 3D shape.
${11~cm^3}$
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2020-10-26 10:39:45
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http://clay6.com/qa/24841/sum-of-the-series-s-large-frac-large-frac-large-frac-large-frac-infty-is-
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Browse Questions
# Sum of the series $\;S=\large\frac{4}{7}-\large\frac{5}{7^2}+\large\frac{4}{7^3}-\large\frac{5}{7^4}+.....\;\infty\;$ is :
$(a)\;\large\frac{4}{43}\qquad(b)\;\large\frac{3}{11}\qquad(c)\;\large\frac{23}{48}\qquad(d)\;\large\frac{23}{24}$
Can you answer this question?
Answer : (c) $\;\large\frac{23}{48}$
Explanation : $S=\large\frac{4}{7}-\large\frac{5}{7^2}+\large\frac{4}{7^3}-\large\frac{5}{7^4}+....\;\infty$
Explanation : $S=\large\frac{4}{7}-\large\frac{5}{7^2}+\large\frac{4}{7^3}-\large\frac{5}{7^4}+....\;\infty$
Explanation : $\large\frac{1}{7}S=\large\frac{4}{7^2}-\large\frac{5}{7^3}+\large\frac{4}{7^4}+....\;\infty$
$\large\frac{8}{7}S=\large\frac{4}{7}-\large\frac{1}{7^2}-\large\frac{1}{7^3}-\large\frac{1}{7^4}+...\;\infty$
$\large\frac{8}{7}S=\large\frac{4}{7}-\;[\large\frac{\large\frac{1}{7^2}}{1-\large\frac{1}{7}}]$
$\large\frac{8}{7}S=\large\frac{4}{7}-\large\frac{1}{4^2}$
$8S=4-\large\frac{1}{6}$
$S=\large\frac{23}{48}\;.$
answered Jan 23, 2014 by
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2017-05-25 12:34:00
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https://www.mathimatikoi.org/forum/viewtopic.php?f=27&t=1227&view=print
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Page 1 of 1
Dobiński’s formula
Posted: Sat Aug 12, 2017 4:29 pm
Let $n \in \mathbb{N}$ and $\mathcal{B}_n$ denote the $n$ - th Bell number. Prove that
$$\sum_{k=0}^{\infty} \frac{k^n}{k!}=\mathcal{B}_n \cdot e$$
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2020-09-26 19:34:03
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http://www.pogonina.com/index.php?option=com_content&task=view&id=518&lang=russian
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4. () e-mail -. Java-script 21:58 11 2010 . Opening tree? I prefer storing analysis in a tree and use Aquarium for that. There you can colour moves with the keys 2-5 (in green, blue, black, orange, red) and I quickly can enter the lines I'm analysing. It works with Chessbase and their tree format, too, but I switched to Aquarium since editing is much easier there. If your analysis is stored as game files, you can add them to a tree, too. Another thing is "IDeA" in Aquarium, I don't know if you've heard about it, but I know some IMs being big fans of it It creates analysis trees with engine evaluations, this is also quite useful for exploring new lines.
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2015-07-04 16:00:46
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http://lxat.quellidelparco.it/benzyl-bromide-sn1-or-sn2.html
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Chloromethylalkyl ethers or sulphides are also reagents for the quaternization of the pyridine nitrogen. org are unblocked. Neophyl bromide and n-butyl bromide are both primary halides. Caution Wear disposable gloves and avoid skin contact. Lesser known is the neopentyl bromide, which is a primary substrate so it should react quickly via SN2, but it does not. The books include the 5th edition of March’s Advanced Organic Chemistry (Wiley), volumes 6–11 of the Compendium of Organic Synthetic Methods (Wiley), Organic Chemistry a Two Semester Course (HarperCollins) into its 2nd edition, and Organic Synthesis (McGraw-Hill) through its 2nd edition. 3) Allyl bromide is a primary alkyl halide, yet it reacts rapidly with silver nitrate in ethanol. In the second case, the reactions corresponds to a SN2, so the less sterically hindered C-next to the Br will be the fastest reaction: 1-bromopropane>benzyl bromide> 2-bromopropane>allyl bromide>2-bromo-2-methylpropane>bromobenzene. The chloride groups go out creating a heterolyic rupture of the compound. The reaction of tert-butyl bromide, (CH3)3CBr, with methanol in an inert solvent proceeds by an SN1 mechanism to give tert-butyl methyl ether, (CH3)3COCH3. Q: Furfuryl chloride can undergo substitution by both SN2 and SN1 mechanisms. The Williamson ether synthesis is an S N 2 reaction in which an alkoxide ion is a nucleophile that displaces a halide ion from an alkyl halide to give an ether. PubChem Substance ID 24892543. In chemistry, an alcohol is any organic compound in which the hydroxyl functional group (–OH) is bound to a carbon. Benzyl bromide (32. The reaction goes in two steps by way of carbocation intermediate. You could change the solvent to something polar aprotic like CH3CN or DMSO and you could use a better base for a nucleophile such as NH 2 - or OH-. Detailed kinetic and product studies have been made on solvolysis of α-(pentamethyldisilanyl)benzyl halides (1a-X; X=Cl and Br) and 1,1,2,2-tetramethyl-1,2-disilaindan-3-yl chloride (2-Cl) in various solvents. Silver nitrate solution can be used to find out which halogen is present in a suspected halogenoalkane. 1016/S0040-4039(00)75465-7, PMID 5701078. The mass spectrum of benzyl fluoride A2. For an Sn2 reaction, the primary halides are most reactive (except for methyl bromide). For Sn1 reactions, an intermediate carbocation must. Aldehydes and ketones react with primary amines to give a reaction product (a carbinolamine) that dehydrates to yield aldimines and ketimines (Schiff bases). Assertion : Benzyl bromide when kept in acetone water it produces benzyl alcohol. fication reaction between potassium p-nitrobenzoate and benzyl bromide catalysed by various crown ethers in a solid-chloroform phase transfer s. (b) Reaction (1) because water is a more polar solvent than methanol, and S N1 reactions take place faster in more polar solvents. If there is no reaction after 5 min, write, "does not react" in your notebook. Benzyl chloride is an alkyl (not aromatic) halide. If there is noreaction after 5 min, write, “does not react” in your notebook. Start learning today!. For an Sn2 reaction, the primary halides are most reactive (except for methyl bromide). 23P: Choose the member of each pair that will react faster by the SN1 me 6. Both of these effects inhibit nucleophilic substitution reactions of either the SN1 or SN2 type, thus net reactivity of the molecule is considerably less than that of saturated alkyl halides. The reaction of tert-butyl bromide, (CH3)3CBr, with methanol in an inert solvent proceeds by an SN1 mechanism to give tert-butyl methyl ether, (CH3)3COCH3. 6 The Williamson Ether synthesis. CH2-Br CH2=CH-CH2-Br benzyl bromide allyl bromide 65. bromocyclopentane (CH2)5Br - Secondary alkyl halide: Sn1 or Sn2 depending on conditions. B) Benzyl bromide hydrolysis done clear. In many cases, at a suitable temperature, it is a mix of the two. In fact, due to these three reasons, the probability of the SN2 pathway occurring is almost negligible in I, II, and IV. The reader is strongly encouraged to review the pages on S N 2 and E2 reactions along with this page. 52) Any involvement of solvent in the reaction cannot be detected in the rate law because the con-centration of the solvent cannot be changed. 7X10-3 atm-cu m/mole(SRC) derived from its vapor pressure, 42. sodium iodide is a catalyst). (b) Chloromethane. For an Sn2 reaction, the primary halides are most reactive (except for methyl bromide). 25P: Give the SN1 mechanism for the formation of 2-ethoxy-3-methylbutane. the S N Ar (addition-elimination) mechanism; the aromatic S N 1 mechanism encountered with diazonium salts. 37 High reactivity of allyl and benzyl halides On the other hand, allyl halides and benzyl halides are more reactive than saturated halides. 1) a slow ionisation of t-butyl bromide to form the ions (CH3)3C+ and Br- 2) a fast attack by OH- to form t-butyl alcohol (CH3)3COH The slow step is always the one that determines the overall rate of reaction, and you can see that this first step involves only one reactant molecule i. All of the rules of the SN1 and SN2 mechanisms apply (refer to those entries under "Reactions of Alkyl Halides"). bromopropene), benzyl bromide, and bromobenzene. hydrolysis of serylglycine. It is a tertiary alkyl halide. The Organic Chemistry Tutor 11,001 views. Halogen containing organic compounds are relatively rare in terrestrial plants and animals. ) Determine the structure of a. eg: The lightest benzylic carbocation 1 is called the benzyl carbocation. CH2Br CH3CH2OH, heat CH2OCH2CH3 => Chapter 17 50 SN2 Reactions • Benzylic halides are 100 times more reactive than primary halides via SN2. B) Benzyl bromide hydrolysis done clear. C) Reaction of NaOH with dinitrofluorobenzene done clear. CH3Br + OH– -> CH3OH + Br– 2. Chlorination of Alkanes. Metal-catalyzed enantioselective allylation, which involves the substitution of allylic metal intermediates with a diverse range of different nucleophiles or SN2'-type allylic substitution, leads to the formation of C-H, -C, -O, -N, -S, and other bonds with very high levels of asym. View Solution play_arrow. An allylic system has a minimum of 3 carbons. The SN1 reaction proceeds stepwise. 1) a slow ionisation of t-butyl bromide to form the ions (CH3)3C+ and Br- 2) a fast attack by OH- to form t-butyl alcohol (CH3)3COH The slow step is always the one that determines the overall rate of reaction, and you can see that this first step involves only one reactant molecule i. Li, a medicinal chemist, summarizes name reactions relevant to heterocyclic chemistry. SN1 versus SN2 Reactions. Why are allyl bromide and benzyl bromide very reactive in both SN1 and Sn2 reactions. This supported an SN1-like carbocation for this process. CHCl 3 (Chloroform: organic solvent) CF 2 Cl 2 (Freon-12: refrigerant CFC) CF 3 CHClBr (Halothane: anesthetic) Halogen atoms are more electronegative than carbon atoms, and so the C-Hal bond is polarized. It is polar covalent molecule with chemical formula, HBr. After removing the cooling bath, stirring continued for 10 min at room temperature and then benzyl bromide (1. Because SN1 reaction goes through carbonation it goes well with Benzyl bromide that has a good , stabilized carbocation As for the SN2 reaction , the primary halides are the most reactive hope this helps. CH4 + Cl2 --(hv)--> CH3Cl + HCl SN2 works well, E2 with KOtBu SN1 and E1 don't work secondary (2°) R2CHX : SN2 works with a good nucleophile E2 works with KOtBu SN1 and E1 occur without strong base or nucleophile tertiary (3°) R3CX : SN1 works well with a good nucleophile. Because SN1 reaction goes through carbonation it goes well with Benzyl bromide that has a good , stabilized carbocation As for the SN2 reaction , the primary halides are the most reactive hope this helps. Assertion : Benzyl bromide when kept in acetone water it produces benzyl alcohol. CH2-Br CH2=CH-CH2-Br benzyl bromide allyl bromide 65. Both of these are primary halides and will be reactive. C) 1-bromo-1-butene. 3 Effective date : 01. Below we see the condensation between ethanol and benzyl bromide: It is important to note that this reaction does not work well at all if secondary or tertiary alkyl halides are used: remember that the alkoxide ion is a strong base as well as a nucleophile, and elimination will compete with nucleophilic substitution (Section 8. Halogen containing organic compounds are relatively rare in terrestrial plants and animals. Other features of the S N 2 mechanism are inversion at the. Since the introduction of ionic liquids (IL) in 1914, there has been a growing interest in this class of materials, because of their unique properties…. Reason : The reaction follows S N 2 mechanism. Why do you think bromobenzene does not react under either substitution conditions? 5. Reactions that make two molecules from one are especially helped by increasing temperature. Study Quiz 11 flashcards from alcohol from the reaction of tert-butyl bromide (t-BuBr) with water by an SN1 mechanism? in a SN2 reaction. 1 Michael/SN2 Reactions with the Halide on the Donor 723 25. 37 High reactivity of allyl and benzyl halides On the other hand, allyl halides and benzyl halides are more reactive than saturated halides. Nucleophilic tendencies of thiophenoxide, azide, chloride, and ethoxide ions toward a series of alkyl and benzyl bromides, as well as substituent effects on rates of SN2 reactions of benzyl. So we form benzyl alcohol for this reaction. It is polar covalent molecule with chemical formula, HBr. 3-Bromocylcohexene is a secondary halide, and benzyl bromide is a primary halide. Then reduce benzylazide with catalytic hydrogenation (H 2, Pd) to form benzylamine (Ph-CH 2-NH 2). sodium iodide is a catalyst). Why does benzyl chloride react under both SN1 and SN2 conditions? A ) Benzyl chloride reacts via both S N 1 and S N 2 conditions because it is a primary substrate hence favors S N 1 but the carbocation also forms at a benzylic position therefore the pi bonds in the ring can stabilize the charge. vstem at 25°C [Reaction 2. Stereo Chemistry Ppt - authorSTREAM Presentation. 6323–6326, doi: 10. If two substrates have very similar reactivity, you can identify them as. A nucleophilic substitution is a substitution reaction in organic chemistry in which the nucleophile displaces a good leaving group, such as a halide, on an aromatic ring. For an Sn2 reaction, the primary halides are most reactive (except for methyl bromide). If both assertion and reason are true and the reason is the correct explanation of the assertion. CHM 331 > Quiz 11 > Flashcards What is the equation for the rate of formation of tert-butyl alcohol from the reaction of tert-butyl bromide (t-BuBr) with water by an SN1 mechanism? Identify the halide(s) that react in a SN2 reaction. It is a colorless liquid with lachrymatory properties. However, the nature of the solvent does play a critical role in this reaction. B) bromobenzene. SN2 is a kind of nucleophilic substitution reaction mechanism. Benzyl bromide is a primary alkyl halide. 3 Effective date : 01. Illustrate the electron flow with curved arrows. Both of these are primary halides and will be reactive. Both of these effects inhibit nucleophilic substitution reactions of either the SN1 or SN2 type, thus net reactivity of the molecule is considerably less than that of saturated alkyl halides. PROBLEM 6-26: Propose a mechanism involving a hydride shift or an alkyl shift for. What is Chemistry? Chemistry is a subdiscipline of science that deals with the study of matter and the substances that constitute it. If there is noreaction after 5 min, write, “does not react” in your notebook. The S N 2 reaction is a type of reaction mechanism that is common in organic chemistry. Sn reaction 1. The reaction proceeds through a substitution bimolecular (SN2) mechanism where the alkoxide anion displaces the halogen of the primary alkyl halide to form an ether. Aldehydes and ketones react with primary amines to give a reaction product (a carbinolamine) that dehydrates to yield aldimines and ketimines (Schiff bases). E) 2-bromo-2. The resulting solution was stirred for 1 h, then quenched by the slow addition of water (70 mL) and extracted with EtOAc (50 mL) several times. Ro;lf:! R. Tuy nhiên, iốt là đắt tiền và một cách giải quyết vấn đề là sử dụng một lượng iốt xúc tác. that substitutions at the benzylic position can be either SN1 or SN2 but the german Wikipedia article on nucleophilic Substitution states that SN2 reactions can never occur there. 5 All the hydrogens in dimethyl ether (CH3OCH3) are bonded to carbon; therefore, intermolecular hydrogen bonding between dimethyl ether molecules does not take place, and its boiling point is lower than that of ethanol (CH3CH2OH), where hydrogen bonding involving the @OH group is important. This is the most common type of hydrolysis. Bromobenzene does not react via SN1 or SN2 pathway because the structure of the ring does not allow for a backside attack in the case of SN2 or the formation of a carbocation in SN1. The cis and trans isomers of 1-bromo-4-tert-butylcyclohexane react at different rates with KOC(CH 3) 3 to yield the same mixture of enantiomers A and B. Hydrobromic acid,Hydrobromic acid chemical formula,properties of hydrobromic acid,Hydrobromic acid pH calculation,Hydrobromic acid uses. Since two reacting species are involved in the slow (rate-determining) step, this leads to the term substitution nucleophilic (bi-molecular) or S N 2. 5 kg in glass bottle 25, 100, 500 g in glass bottle Safety & Documentation. University of Illinois UIC at Chicago CHEM 232, Spring 2010 Slide Lecture 26: April 15 Allylic SN2 Faster: Two Arguments 2. And we end up with an OH replacing our Br. Search Search. Neophyl bromide and n-butyl bromide are both primary halides. S N2 backside attack is easy on the methyl group. The method of conductivity is used to measure the degree of dissociation of alkali halide in acetone. Once the bromide left with an electron pair, the nucleophile replaced it and the result was a product of benzyl ether. Dear Organic Chemistry Student, Are you struggling For example, say the electrophile was benzyl bromide - it could do SN1 (resonance stabilized carbocation) or SN2 (primary alpha carbon) and the nucleophile was ethoxide, which could do SN2 (strong nucleophile) or E2 (strong base). SN1 reactions can be preparatively useful in organic synthesis, but only in cases where: Particularly stable carbocations are formed, and elimination reactions are either impossible, or reactions conditions have been adjusted in such a way that elimination reactions are suppressed. If you're seeing this message, it means we're having trouble loading external resources on our website. SN1 Mechanism - Evidence. Presentations (PPT, KEY, PDF). Chapter 17 49 SN1 Reactions • Benzylic carbocations are resonancestabilized, easily formed. OH OH2+ Br. Question: These are two questions from the Nucleophilic Substitution Lab in Organic Chemistry: 1) Benzyl bromide is a primary alkyl halide that reacts rapidly with sodium iodide in acetone, but it. It is a tertiary alkyl halide. In these cases the reactions proceed via the SN1 mechanism9-11. Detailed kinetic and product studies have been made on solvolysis of α-(pentamethyldisilanyl)benzyl halides (1a-X; X=Cl and Br) and 1,1,2,2-tetramethyl-1,2-disilaindan-3-yl chloride (2-Cl) in various solvents. As generally defined herein, R sc is optionally substituted C 1-6 alkyl, —OR SO, —C(═O)N(R SN1) 2, or —N(R SN2) 2, wherein R SN1 and R SN2 are as defined herein. If a more polar solvent with a Lewis acid is used, it will react SN1. Assertion : Benzyl bromide when kept in acetone water it produces benzyl alcohol. Bromocyclohexane on the other hand should have formed a precipitate with either SN1 or SN2 since it is a secondary halide, but we did not observe a reaction, so perhaps a contamination occurred. 86) Idenitfy the halide(s) that react in a SN1 reaction. 7k points). E) 2-bromo-2. It also sought to find. CHCI Benzyl Chloride 7 CH2=CHCH2Br Allyl Bromide 8 PhBr Bromobenzene. Phenyl groups are closely related to benzene and can be viewed as a benzene ring, minus a hydrogen, which may be replaced by some other element or compound to serve as a functional group. 7X10-3 atm-cu m/mole(SRC) derived from its vapor pressure, 42. That makes it easier for the halogen to leave as an anion and form silver or some other halide. SN2 is a kind of nucleophilic substitution reaction mechanism. This has bimolecular kinetics: rate = k [CH3CH2I] [CN-] SN1 stands for substition nucleophillic unimolecular. So we form benzyl alcohol for this reaction. The influence of the pyridone/benzyl bromide ratio was studied. B) Benzyl bromide hydrolysis done clear. The cis and trans isomers of 1-bromo-4-tert-butylcyclohexane react at different rates with KOC(CH 3) 3 to yield the same mixture of enantiomers A and B. Organic Chemistry I-Lab (CHEM 211) Uploaded by. Start learning today!. CHEM 51LB Lecture Notes - Waste Container, Benzyl Bromide, Fume Hood Lecture Note CHEM 51LB Study Guide - Final Guide: Abstract Window Toolkit, Asteroid Family, Aldehyde. The resulting solution was stirred for 1 h, then quenched by the slow addition of water (70 mL) and extracted with EtOAc (50 mL) several times. Journal of Physical Organic Chemistry 1994 , 7 (5) , 234-243. org are unblocked. Explain these observations. 6 M solution in ethanol 36/37/38 26-39 מידע נוסף. Since two reacting species are involved in the slow (rate-determining) step, this leads to the term substitution nucleophilic (bi-molecular) or S N 2. As described in the previous section, a majority of the reactions thus far described appear to proceed by a common single-step mechanism. 3-Phenylbutyl benzyl ether can. Benzylic groups are especially activated for nucleophilic substitution due to the facile formation of benzylic cations in the case of SN1 processes or for SN2 processes, by the overlap of the transition state quasi p-orbital with the π-system of the arene. 24P: 3-Bromocyclohexene is a secondary halide, and benzyl bromide is a p 6. Lesser known is the neopentyl bromide, which is a primary substrate so it should react quickly via SN2, but it does not. Nucleophilic Substitution: SN1 vs SN2 Factors SN1 SN2 Rate Law unimolecular bimolecular Number of Steps 2 1 Strength of Nucleophile Weak (e. Why are allyl bromide and benzyl bromide very reactive in Answers. (b) Chloromethane. The tosylate leaves to generate a carbocation. Substitution reaction, any of a class of chemical reactions in which an atom, ion, or group of atoms or ions in a molecule is replaced by another atom, ion, or group. R groups that make more stable carbocations react faster 3° > 2° > 1° > CH3 tertiary RX react by SN1 CH3 and primary RX react by SN2 secondary RX react either way SN1 Mechanism - X Groups. 22 (a) Reaction (2) because bromide ion is a better leaving group than chloride ion. The reaction of tert-butyl bromide, (CH3)3CBr, with methanol in an inert solvent proceeds by an SN1 mechanism to give tert-butyl methyl ether, (CH3)3COCH3. Aryl halides are unreactive to S N2 conditions. For the SN1 reaction you will use a solution of silver nitrate in ethanol. The final products are typically alkyl halides, unless one of the carbons is part of an aromatic ring which would preclude reaction at that center. Interactive 3D animation of SN2 substitution at a benzylic centre for students studying advanced school chemistry and University chemistry. Due to having the resonance effect the benzyl carbocation is the considered to be stable causing it to undergoes the Sn1 reaction easily. 5 All the hydrogens in dimethyl ether (CH3OCH3) are bonded to carbon; therefore, intermolecular hydrogen bonding between dimethyl ether molecules does not take place, and its boiling point is lower than that of ethanol (CH3CH2OH), where hydrogen bonding involving the @OH group is important. (CH3)3CBr ( Heat at 55 C in EtOH/solvolysis) = (CH3)3COEt 72%SN1 + 28% CH3CCH3CH2 E1 + HBr. Br is a better leaving group than Cl,so the arrangement would be : n-butyl chloride > sec-butyl chloride >tert-butyl chloride. CHEM 51LB Lecture Notes - Waste Container, Benzyl Bromide, Fume Hood Lecture Note CHEM 51LB Study Guide - Final Guide: Abstract Window Toolkit, Asteroid Family, Aldehyde. why is benzyl bromide which appears to be a primary halide able to undergo sn2 and sn1 reactions support your answers Q : Discuss why adding nai increases the reaction rate If sodium iodide is added to the reaction mixture, the rate of methanol formation is dramatically increased (i. Stephen Daniel of QuantumTechniques. 1 Michael/SN2 Reactions with the Halide on the Donor 723 25. Router Screenshots for the Sagemcom Fast 5260 - Charter. Organic Chemistry I-Lab (CHEM 211) Uploaded by. Since the introduction of ionic liquids (IL) in 1914, there has been a growing interest in this class of materials, because of their unique properties…. The method of conductivity is used to measure the degree of dissociation of alkali halide in acetone. Presentations (PPT, KEY, PDF). Metal-catalyzed enantioselective allylation, which involves the substitution of allylic metal intermediates with a diverse range of different nucleophiles or SN2'-type allylic substitution, leads to the formation of C-H, -C, -O, -N, -S, and other bonds with very high levels of asym. The formation of the sigma complex is an endothermic and energetically unfavorable process - it is therefore the. Allylic Carbocation. 3-Bromocylcohexene is a secondary halide, and benzyl bromide is a primary halide. George Hademenos - Schaums Outline of Theory and Problems of Organic Chemistry (1999 McGraw-Hill). Reaction proceeds via SN1 because a tertiary carbocation was formed, the solvent is polar protic and Br- is a good leaving group. Was there any observable rate difference between these two substrates under SN2 conditions? Explain your result. Factors Affecting the Relative rates of Nucleophilic Substitution Reactions. Rvlf:! Sw1 fast. Benzyl bromide is an organic compound with the formula C 6 H 5 CH 2 Br. " problems retrosynthesis sets retrosynthetic analysis rules of thumb Self-directed Learning sigma framework SmartWork Smartwork5 SN1/SN2 SN1/SN2/E1/E2 Solubility Solved Problems solvents Spectra Spectroscopy. 69X10+2 mg/L(2). It should undergo E2 (tertiary alkyl halide and strong base), but there are no beta hydrogens available. 16) Sn1 - secondary substrate plus weak nucleophile. CH3CH2CH2CH2Br + CH3Br. • Benzyl halides undergo SN1 reactions. Chapter 9 - Discussion Answers. Request for Solution File. 1016/S0040-4039(00)75465-7, PMID 5701078. (a) bromobenzene or benzyl bromide? (b) CH 3Cl or (CH 3) 3CCl (c) CH 3CH=CHBr or H 2C=CHCH 2Br (a) Benzyl bromide. Fill In The Table 2 According To The Best Of Your Ability. Also remember both are secondary carbons making them even less favorable for sn2. Acid: Water can act as an acid or a base, according to the Bronsted-Lowry acid theory. The solution temperature (0o C) greatly slows Sn1 carbocation formation. Bromobenzene does not react via SN1 or SN2 pathway because the structure of the ring does not allow for a backside attack in the case of SN2 or the formation of a carbocation in SN1. EC Number 201-151-7. It is introduced into the aromatic compounds and acts as intermediate for the preparation of pharmaceuticals, agrochemicals and other kinds of chemicals. What about benzyl bromide? Write a mechanism showing its reaction with water. 1 Michael/SN2 Reactions with the Halide on the Donor 723 25. Both of these are primary halides and will be reactive. Br (n-butyl bromide) and (CH 3) 2 CHCH 2 Br (iso-butyl bromide). Reaction with acid halides. Hydrobromic acid,Hydrobromic acid chemical formula,properties of hydrobromic acid,Hydrobromic acid pH calculation,Hydrobromic acid uses. SN2 reaction of the amine with an excess of CH3I in the first step yields an. 23P: Choose the member of each pair that will react faster by the SN1 me 6. ++CH2 CH2 etc ++CH2 CH2 etc 66. A) benzyl bromide. Draw a mechanisms for all the types of reactions we have done in lab so far (sn1, sn2, bromination and oxidation) Determine what type of reaction will occur, given reagents, but no products. Chemistry 2283g Experiment 1 – Alkyl Halides EXPERIMENT 1: Preparation and Reactivity of Alkyl Halides th Relevant Sections in the text (Wade, 7 ed. Based on this Henry's Law constant, the volatilization half-life. Look it up in the book, it is the chapter regarding phenols. if a LG is good for SN2, it is good for SN1. Nucleophilic Substitution Reactions are one of the most important major classes of organic chemistry and essential that you build a solid foundation and understanding of their principles and mechanisms such as the SN1 and SN2 reactions. จงอ านชื่อสารต อไปนี้แบบ IUPAC 1) H3C CHCH2CHCH3 Cl CH3 2 Chloro 4 methylpentane 2) CH2CH2CHCH2Cl Cl Cl 1,2,4 Trichlorobutane Br. H2O, ROH) Strong (e. The reaction mixture was then cooled to RT and filtered subsequently to remove insoluble salts, which were washed further with EtOAc. This has bimolecular kinetics: rate = k [CH3CH2I] [CN-] SN1 stands for substition nucleophillic unimolecular. This was then converted as before into the enantiopure (aR) trichloroacetimidate 25. 2 Answers to a) Write the chemical equation for all the reactions both SN1 and SN2 involving primary and secondary substrates b) Rank the primary halide in order of increasing reactivity toward SN2 c) Rank the tietiary substrate in order of increasing reactivity toward SN1 - 120297. However, the halogen rich environment of the ocean has produced many interesting natural products incorporating large. Write the potential sn2 and sn1 reaction as well as the potential the sn2 and sn1 mechanism. Resonance distributes the positive charge. Also find the composition of the vapour phase. in the first one the reaction would be SN2, so the best reactive alkyl halide would be the one with the best leaving group,and connected to the type 1 carbon,which is desirable for SN2. CHEM 51LB Lecture Notes - Waste Container, Benzyl Bromide, Fume Hood Lecture Note CHEM 51LB Study Guide - Final Guide: Abstract Window Toolkit, Asteroid Family, Aldehyde. For the SN2 reaction you will use a solution of sodium iodide in 2-butanone. CH3CH2CH2CH2Br > (CH3)2CHCH2Br > CH3CH2CH(Br)CH3 > (CH3)3CBr (SN2). Aldehydes and ketones react with primary amines to give a reaction product (a carbinolamine) that dehydrates to yield aldimines and ketimines (Schiff bases). However, in this experiment the DABCO will always be in excess so further reaction is unlikely. How can this be explained if there is no carbocation intermediate? 1. Look it up in the book, it is the chapter regarding phenols. R in Acetic Acid 33%; HydroBromic Acid 48%; Methyl Bromo Acetate; Propionyl Bromide; Organic Bromide. Since T-Butyl Chloride is an alkylating agent for SN1 reactions, it can be used a protecting group for alcohols and can be de-protected using the acidic conditions. Nucleophilic tendencies of thiophenoxide, azide, chloride, and ethoxide ions toward a series of alkyl and benzyl bromides, as well as substituent effects on rates of SN2 reactions of benzyl. Q: Furfuryl chloride can undergo substitution by both SN2 and SN1 mechanisms. 16 g/ml: boiling point. The electrochemical fixation of CO2 into organic substrates is a convenient method of synthesis of carboxylic acids I. Also remember both are secondary carbons making them even less favorable for sn2. Since the introduction of ionic liquids (IL) in 1914, there has been a growing interest in this class of materials, because of their unique properties…. Compound A Compound B b. It is polar covalent molecule with chemical formula, HBr. tertiary RX react by SN1 CH3 and primary RX react by SN2 E2 works with KOtBu SN1 and E1 occur without strong base or. That makes it easier for the halogen to leave as an anion and form silver or some other halide. Write the mechanism of the following reaction: nBuBr + KCN $$\underset{}{\stackrel{EtOH - H_2O}{\longrightarrow}}$$ nBuCN. They found that the S 2 rate constants N catalyzed by various crown ethers decreased in the order: oxydimethylene-bis-benzo-15-crown-5 dicyclohexyl-18-crown-6 18-crown-6. CHCI Benzyl Chloride 7 CH2=CHCH2Br Allyl Bromide 8 PhBr Bromobenzene. If two substrates have very similar reactivity, you can identify them as. Fill In The Table 2 According To The Best Of Your Ability. You could change the solvent to something polar aprotic like CH3CN or DMSO and you could use a better base for a nucleophile such as NH 2 - or OH-. SN2 reactions are substitution nucleophilic bimolecular. There are two mechanisms for nucleophilic. Both reactants could do SN2, so that will be the major reaction pathway. If you're behind a web filter, please make sure that the domains *. David Rawn, in Organic Chemistry Study Guide, 2015. Allylic and Benzylic compounds Allylic and benzylic compounds are especially reactive in SN2 reactions. S N 2 is a kind of nucleophilic substitution reaction mechanism. In: Tetrahedron Lett. Both of these are primary halides and will be reactive. Summary of Sn1 and Sn2 reactions and the types of molecules and solvents that favor each. (a) CH3CH2CH2CH2Br < (CH3)2CHCH2Br < CH3CH2CH(Br)CH3 < (CH3)3CBr (SN1). Hydrobromic acid is an important chemical in chemistry, especially in inorganic chemistry. Testing for halogenoalkanes. The Organic Chemistry Tutor 11,001 views. The resulting solution was stirred for 1 h, then quenched by the slow addition of water (70 mL) and extracted with EtOAc (50 mL) several times. Write the mechanism of the following reaction: nBuBr + KCN $$\underset{}{\stackrel{EtOH - H_2O}{\longrightarrow}}$$ nBuCN. E) 2-bromo-2. D) 1-bromo-2-butene. ELIGIBILITY FOR THE AWARD OF DEGREE. SN2 is a kind of nucleophilic substitution reaction mechanism. C) Reaction of NaOH with dinitrofluorobenzene done clear. The method of conductivity is used to measure the degree of dissociation of alkali halide in acetone. Neophyl bromide and n-butyl bromide are both primary halides. Introduction to the Chemistry of Alkyl Halides Introduction An alkyl halide is another name for a halogen-substituted The polarity makes the C atom electrophilic and prone to attack by nucleophiles via SN1 or SN2 reactions. Path A represents most probably an SN2 type displacement (carbenium ions implied by an SN1 reaction would not be stabilized by electron-deficient phenyl rings). S N 2 is a kind of nucleophilic substitution reaction mechanism. Benzylic halides undergo nucleophilic substitution reactions very readily (review) 1obenzylic halides typically react via an SN2 pathway (review), and there is no competition from elimination. Benzyl bromide is a primary alkyl halide. Bromobenzene does not react via SN1 or SN2 pathway because the structure of the ring does not allow for a backside attack in the case of SN2 or the formation of a carbocation in SN1. It is polar covalent molecule with chemical formula, HBr. Explain this seemingly anomalous result. N-Alkylation through mechanism A (SN2 mechanism) is kinetically favored while C-alkylation through an SN1-type mechanism is thermodynamically favored and is obsd. The method of conductivity is used to measure the degree of dissociation of alkali halide in acetone. Packaging 2. Give an example of both an SN1 and SN2 mechanism. 2 Answers to a) Write the chemical equation for all the reactions both SN1 and SN2 involving primary and secondary substrates b) Rank the primary halide in order of increasing reactivity toward SN2 c) Rank the tietiary substrate in order of increasing reactivity toward SN1 - 120297. Using your experimental data, rank the halides in table 1 by their SN2 and SN1 reactivity. Bromobenzene doesn’t react under either of these conditions. benzyl bromide. Due to having the resonance effect the benzyl carbocation is the considered to be stable causing it to undergoes the Sn1 reaction easily. 37 High reactivity of allyl and benzyl halides On the other hand, allyl halides and benzyl halides are more reactive than saturated halides. The tosylate leaves to generate a carbocation. 1) a slow ionisation of t-butyl bromide to form the ions (CH3)3C+ and Br- 2) a fast attack by OH- to form t-butyl alcohol (CH3)3COH The slow step is always the one that determines the overall rate of reaction, and you can see that this first step involves only one reactant molecule i. Both of these are primary halides and will be reactive. Moreover, the primary alkyl halides would undergo SN2, the secondary alkyl halides could undergo either SN1 or SN2 (esp. Li, a medicinal chemist, summarizes name reactions relevant to heterocyclic chemistry. if a LG is good for SN2, it is good for SN1. Safety Data Sheet according to 29CFR1910/1200 and GHS Rev. B) bromobenzene. Sn2 for the different halides used in this experiment. 5 kg in glass bottle 25, 100, 500 g in glass bottle Safety & Documentation. Why benzyl chloride is highly reactive in SN1 reaction in spite of primary alkyl halide ? How the rate of SN1 and SN2 reaction depends on the nature of leaving group ? Which one is more active between 1-iodo butane and 1-chloro butane towards SN2 reaction ? Why vinyl chloride is inactive in SN2 reaction ? Why chloro benzene is inactive in SN2 reaction ?. The mass spectrum of 2-benzyl phenol A4. The explanation of rearrangement sounds good in theory, but this is incorrect as a primary carbocation would have to be formed before rearrangement. Path A represents most probably an SN2 type displacement (carbenium ions implied by an SN1 reaction would not be stabilized by electron-deficient phenyl rings). vstem at 25°C [Reaction 2. จงอ านชื่อสารต อไปนี้แบบ IUPAC 1) H3C CHCH2CHCH3 Cl CH3 2 Chloro 4 methylpentane 2) CH2CH2CHCH2Cl Cl Cl 1,2,4 Trichlorobutane Br. ) Determine the structure of a. What determines whether 2-bromobutane undergoes SN1 and/or SN2 reactions? * 2. 22P: Propose an SN1 mechanism for the solvolysis of 3-bromo-2,3-dimethyl 6. 24P: 3-Bromocyclohexene is a secondary halide, and benzyl bromide is a p 6. In organic chemistry, the phenyl group or phenyl ring is a cyclic group of atoms with the formula C 6 H 5. A) benzyl bromide. What is the role of the silver inthe sn1 reactions? Why is it necessary? 4. D) Sulphonation of benzene done clear. semester (s2) assignment unique number: 359125 question due date: 30 august 2013 before doing this assignment study the following: study units and from the. Toxic by all routes (ie, oral, dermal, inhalation), exposure to benzyl bromide may occur environmentally from inhaling air contaminated by exhaust from automobiles using leaded gasoline, or occupationally from its use in research. why do you think allyl bromide and benzyl bromide undergo both sn1 and sn2 reactions? 2. 塩化ベンジル（えんか—、benzyl chloride）は、有機合成で用いられる芳香族化合物の一種。示性式は C6H5CH2Cl、トルエンのメチル基の水素をひとつ塩素に置き換えた構造を持ち、α-クロロトルエン、クロロメチルベンゼンと呼ぶこともできる。. Explain this seemingly anomalous result. Reaction type: Nucleophilic substitution (S N 1 or S N 2). Why is Benzyl Bromide, which appears to be a primary halide, able to undergo SN2 and SN1 reactions? Support your answers with drawings. Since both the allylic $\mathrm{S_{N}1}$ and $\mathrm{S_{N}2}$ reactions are stabilized, there is a delicate balance between the two pathways. An allylic system has a minimum of 3 carbons. 2) Benzyl bromide reacts rapidly with sodium iodide in acetone, and also reacts rapidly with ethanol and silver acetate. The Organic Chemistry Tutor 11,001 views. Reaction Heating Time 1 Table 1: S2 Reactivity Structure Name CH3CH2CH2CH2CI n-Butyl Chloride CH3CH2CH2CH2Br n-Butyl Bromide 3 CH3CHBYCH2CH3 sec-Butyl Bromide 4 (CH3)2CHCH2BI Isobutyl Bromide 5 (CH3),CCI tert-Butyl Chloride 6 CHCHCI Benzyl Chloride 7 CH2=CHCH2BT Allyl Bromide 8 PhBr Bromobenzene 7. Experiment 7 — Nucleophilic Substitution _____ Pre-lab preparation (1) Textbook Ch 8 covers the SN2 and SN1 mechanisms. Why are allyl bromide and benzyl bromide very reactive in Answers. B) bromobenzene. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Neophyl bromide and n-butyl bromide are both primary halides. This serves to further stabilize it and makes the benzyl cation have the reactivity of a secondary carbocation when it comes to SN1. in the first one the reaction would be SN2, so the best reactive alkyl halide would be the one with the best leaving group,and connected to the type 1 carbon,which is desirable for SN2. ? You need to consider what type of molecules are reactive in each one. The solvolysis of benzyl chloride, a primary alkyl halide, proceeds via the formation of a carbocation intermediate and occurs more rapidly than the solvolysis of the other compounds in this experiment. Chlorination of Alkanes. MDL number MFCD00000871. I also thought that it would be SN2 but the solution said SN1. R in Acetic Acid 33%; HydroBromic Acid 48%; Methyl Bromo Acetate; Propionyl Bromide; Organic Bromide. Although no detailed kinetic studies have been conducted, two key observations are more consistent with an SN1-type mechanism. D) Sulphonation of benzene done clear. Acetyl Bromide; Bromo Acetyl Bromide; Ethyl Bromo Acetate; H. The substrate that can make the most. -OH, -OCH3,. 2-Chlorobutane ≥99% Synonym: sec-Butyl chloride CAS Number 78-86-4. For example, say the electrophile was benzyl bromide - it could do SN1 (resonance stabilized carbocation) or SN2 (primary alpha carbon) and the nucleophile was ethoxide, which could do SN2 (strong nucleophile) or E2 (strong base). SN2 reactions - definition In this mechanism of SN2, one bond is broken and one bond is formed synchronously, i. This experiment is aimed at determining the effects of the structure of the alkyl halides on the rate of SN1 and SN2 reactions. For an Sn2 reaction, the primary halides are most reactive (except for methyl bromide). The reaction goes in two steps by way of carbocation intermediate. Relatively harsh conditions are typically required for generating benzyl ethers from the corresponding alcohol, with the two most popular protocols being (1) the Williamson ether synthesis, an SN2-type reaction between alkali metal alkoxides and benzyl bromide, and (2) coupling using benzyl trichloroacetimidate, which is generally promoted by. due the the presence of the polar protic substance, methanol. It provides examples such as the preparation of phenyl propyl. The reaction could proceed by either the SN1 or the SN2 mechanism. 3-Bromocylcohexene is a secondary halide, and benzyl bromide is a primary halide. Vide Appendix-A. The substrate that can make the most. The leaving group first leaves, whereupon a carbocation The SN2 reaction is favored by polar aprotic solvents – these are solvents such as acetone, DMSO. A benzylic carbocation is a resonance-stabilized carbocation in each of the two equally stable major resonance forms of which the formal charge of +1 is on a benzylic carbon. Aldehydes and ketones react with primary amines to give a reaction product (a carbinolamine) that dehydrates to yield aldimines and ketimines (Schiff bases). The electrochemical fixation of CO2 into organic substrates is a convenient method of synthesis of carboxylic acids I. Allyl bromide is a primary alkyl halide, yet it undergoes rapid reaction with silver nitrate in ethanol. question_answer19) Which represents nucleophilic aromatic substitution reaction [Orissa JEE 2004] A) Reaction of benzene with $C{{l}_{2}}$ in sunlight done clear. If 1μg of 90 Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically. DEGREE COURSEs. When this occurs, water spontaneously ionizes into hydroxide anions and hydronium cations. (a) bromobenzene or benzyl bromide? (b) CH 3Cl or (CH 3) 3CCl (c) CH 3CH=CHBr or H 2C=CHCH 2Br (a) Benzyl bromide. ppt), PDF File (. Li, a medicinal chemist, summarizes name reactions relevant to heterocyclic chemistry. CH3CH2CH2CH2Br + NaOCH3d. 2 Michael/SN2 Reactions with the Halide on the Acceptor 724 25. Q:-The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Question: 6. the rate remains the same B. Once the bromide left with an electron pair, the nucleophile replaced it and the result was a product of benzyl ether. What is the nucleophile in the sn1 reactions performed in our lab? -----is it Ethanol? 3. When this was reacted with isobutyl (R)-lactate an approximately 1:1 mixture of the benzyl-protected lactates 22 was obtained (Scheme 2). It is a tertiary alkyl halide. 3) Allyl bromide is a primary alkyl halide, yet it reacts rapidly with silver nitrate in ethanol. If you're behind a web filter, please make sure that the domains *. CH4 + Cl2 --(hv)--> CH3Cl + HCl SN2 works well, E2 with KOtBu SN1 and E1 don't work secondary (2°) R2CHX : SN2 works with a good nucleophile E2 works with KOtBu SN1 and E1 occur without strong base or nucleophile tertiary (3°) R3CX : SN1 works well with a good nucleophile. Explanation:. PROBLEM 6-26: Propose a mechanism involving a hydride shift or an alkyl shift for. Was there any observable rate difference between these two substrates under SN2 conditions? Explain your result. If you thought about an SN1 type reaction, let's go ahead and think about that. A benzylic carbocation is a resonance-stabilized carbocation in each of the two equally stable major resonance forms of which the formal charge of +1 is on a benzylic carbon. Nucleophilic substitutions on benzyl chloride with such basic nucleophiles proceed via SN2. It is a colorless liquid with lachrymatory properties. Finally, react benzylamine with excess 1-bromobutane (N-butyl bromide, CH 3 CH 2 CH 2 CH 2 Br) to give tri-N-butyl benzyl ammonium bromide. DO NOT confuse an allylic group with a vinyl group. It is impossible on the tertiary. CHCI Benzyl Chloride 7 CH2=CHCH2Br Allyl Bromide 8 PhBr Bromobenzene. benzyl bromide. Dear Organic Chemistry Student, Are you struggling For example, say the electrophile was benzyl bromide - it could do SN1 (resonance stabilized carbocation) or SN2 (primary alpha carbon) and the nucleophile was ethoxide, which could do SN2 (strong nucleophile) or E2 (strong base). Why is bromobenzene non-reactive under both SN1 and SN2 Conditions?. Alkyl halides are a class of compounds where a halogen atom or atoms are bound to an sp3 orbital of an alkyl group. Both of these are primary halides and will be reactive. SN1 reactions can be preparatively useful in organic synthesis, but only in cases where: Particularly stable carbocations are formed, and elimination reactions are either impossible, or reactions conditions have been adjusted in such a way that elimination reactions are suppressed. Section: 9-5. Testing for halogenoalkanes. 28 mol) was added via syringe, and the resulting reaction mixture was warmed to 80° C. What is the nucleophile in the sn1 reactions performed in our lab? -----is it Ethanol? 3. Below we see the condensation between ethanol and benzyl bromide: It is important to note that this reaction does not work well at all if secondary or tertiary alkyl halides are used: remember that the alkoxide ion is a strong base as well as a nucleophile, and elimination will compete with nucleophilic substitution (Section 8. 24P: 3-Bromocyclohexene is a secondary halide, and benzyl bromide is a p 6. Based on this Henry's Law constant, the volatilization half-life. This time the slow step of the reaction only involves one species - the halogenoalkane. Br (n-butyl bromide) and (CH 3) 2 CHCH 2 Br (iso-butyl bromide). Benzyl bromide is a primary alkyl halide. For Sn1 reactions, an intermediate carbocation must. Reaction with acid halides. 7 Nucleophilic Substitution. Do typical calculations related to stoichiometry in a reaction (figuring out limiting reagent, calculating % yield, etc. The above results can be rationalized in one of the three ways outlined in the Scheme. If two substrates have very similar reactivity, you can identify them as. S N2 backside attack is easy on the methyl group. This mechanism, called SN1 (substitution, nuceleophilic, unimolecular in the rate-determining step), is followed when R is a tertiary alkyl group and may also be involved when R is a secondary group or when R can form a resonance-stabilized carbocation, such as an allyl or benzyl ion. Benzyl halides react via SN1 and SN2 with equal probability. It is a colorless liquid with lachrymatory properties. Carbocations are very unstable - its carbon with only 6 electrons around it. Both reactants could do SN2, so that will be the major reaction pathway. (b) Chloromethane. E) 2-bromo-2-phenylpropane. If an SN2 reaction occurs, the iodide will displace the bromide or chloride of the alkyl. 1968, 9 (60), S. If 1μg of 90 Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically. 3-Phenylbutyl benzyl ether can. com shares the many uses of combining DMSO and silver for wound healing and more with humans and animals. Posted by 4 years ago. The books include the 5th edition of March’s Advanced Organic Chemistry (Wiley), volumes 6–11 of the Compendium of Organic Synthetic Methods (Wiley), Organic Chemistry a Two Semester Course (HarperCollins) into its 2nd edition, and Organic Synthesis (McGraw-Hill) through its 2nd edition. Reactions Ionic reactions: Radical reactions:Bond breaking and bond Bond breaking and bond makingmaking take place in a …. Answer: Section: 9-5 85) Identify the halide(s) that react in a SN2 reaction. Journal of Physical Organic Chemistry 1994 , 7 (5) , 234-243. If you want an SN1 reaction, then 2-bromobutane should be used because secondary alkyl halides can react both by SN1 and SN2, although SN2 is still preferable. They are known as SN1 and SN2 reactions. Examples of the six test tubes with reactions can be seen in Figure 1 below. Hydrobromic acid,Hydrobromic acid chemical formula,properties of hydrobromic acid,Hydrobromic acid pH calculation,Hydrobromic acid uses. Why are allyl bromide and benzyl bromide very reactive in Answers. 塩化ベンジル（えんか—、benzyl chloride）は、有機合成で用いられる芳香族化合物の一種。示性式は C6H5CH2Cl、トルエンのメチル基の水素をひとつ塩素に置き換えた構造を持ち、α-クロロトルエン、クロロメチルベンゼンと呼ぶこともできる。. Neophyl bromide and n-butyl bromide are both primary halides. CH4 + Cl2 --(hv)--> CH3Cl + HCl SN2 works well, E2 with KOtBu SN1 and E1 don't work secondary (2°) R2CHX : SN2 works with a good nucleophile E2 works with KOtBu SN1 and E1 occur without strong base or nucleophile tertiary (3°) R3CX : SN1 works well with a good nucleophile. They form resonancestabilized carbocations. D) 1-bromo-2-butene. Testing for halogenoalkanes. The present paper reports on a general and convenient route for the preparation of 3-. electron delocalization lends stability to a molecule with interacting p-orbitals. SN2 reactions are substitution nucleophilic bimolecular. Also, provide the name of the mechanism (SN2, SN1, E2 or E1). BENZYL ( GOOD FOR SN1 )IS ALSO A GOOD SN2 SUBSTRATE primary, but faster than other primary CH2 Br + NaI CH2 I + NaBr I overlap in the activated H. and stirred overnight. The allylic carbon and the nearby double bond. answered Jan 14, 2019 in Haloalkanes and Haloarenes by ramesh ( 82. CH3CH2CH2CH2Br > (CH3)2CHCH2Br > CH3CH2CH(Br)CH3 > (CH3)3CBr (SN2). an SN2 reaction with an alkoxide as nucleophile the alkyl halide should be methyl or 1°. Assertion : Benzyl bromide when kept in acetone water it produces benzyl alcohol. General description Benzyl bromide is an aromatic halide that is mainly used as a protecting agent for protecting hydroxy group of alcohols via O-benzylation reaction. 6323–6326, doi: 10. An allylic system has a minimum of 3 carbons. They found that the S 2 rate constants N catalyzed by various crown ethers decreased in the order: oxydimethylene-bis-benzo-15-crown-5 dicyclohexyl-18-crown-6 18-crown-6. D) Sulphonation of benzene done clear. I would suggest you actually try this on your own as I am pretty sure I am going to give you a quiz on it. An sp 3-hybridized electrophile must have a leaving group (X) in order for the reaction to take place. In particular, non-steroidal anriinflammatory agents such as 2-arylpropanoic. For an Sn2 reaction, the primary halides are most reactive (except for methyl bromide). CH3Br + OH– -> CH3OH + Br– 2. Presentations (PPT, KEY, PDF). For Sn1 reactions, an intermediate carbocation must. the rate decreases by a factor of 2. Using your experimental data, rank the halides in table 1 by their SN2 and SN1 reactivity. Then silver nitrate solution is added. This shows that primary reacts faster because there is a less amount of hindrance and so the nuchleophile can attack easily. SN2 Reactions of Benzylic Halides 1 S N2 of benzylic halides is faster than allylic halides. For an Sn2 reaction, the primary halides are most reactive (except for methyl bromide). Br is a better leaving group than Cl,so the arrangement would be : n-butyl chloride > sec-butyl chloride >tert-butyl chloride. Due to having the resonance effect the benzyl carbocation is the considered to be stable causing it to undergoes the Sn1 reaction easily. Factors Affecting the Relative rates of Nucleophilic Substitution Reactions. VERY IMPORTANT INSTRUCTIONS: Kindly refer the official communication of the University in the. Three Lessons from Student Exams. 2 Answers to a) Write the chemical equation for all the reactions both SN1 and SN2 involving primary and secondary substrates b) Rank the primary halide in order of increasing reactivity toward SN2 c) Rank the tietiary substrate in order of increasing reactivity toward SN1 - 120297. SN1: the benzyl carbocation formed in the rate determing step of the reaction is stable as the positive charge generated is spread over the ring. question_answer19) Which represents nucleophilic aromatic substitution reaction [Orissa JEE 2004] A) Reaction of benzene with $C{{l}_{2}}$ in sunlight done clear. Nucleophilic Substitution Reactions are one of the most important major classes of organic chemistry and essential that you build a solid foundation and understanding of their principles and mechanisms such as the SN1 and SN2 reactions. Benzyl bromide (32. This page covers the mechanistically related reaction types, S N 1 and E1. David Rawn, in Organic Chemistry Study Guide, 2015. 25 M solutions in ethanol 11-22-36/37/38 26-37 benzyl bromide 0. Then state which reaction(s) will occur if any, and why? with NaI/acetone and AgNO3/ethanol 1) t-butyl bromide J) benzyl bromide K) bromobenzene I'm having trouble understanding when to use SN1 or SN2 (or both) 0 Sn1 - Sn2. Neophyl bromide and n-butyl bromide are both primary halides. This has unimolecular kinetics: rate = k [Ph2CBr] Review. The silver nitrate test is sensitive enough to detect fairly small concentrations of halide ions. This time the slow step of the reaction only involves one species - the halogenoalkane. Hydrobromic acid is an important chemical in chemistry, especially in inorganic chemistry. CHCI Benzyl Chloride 7 CH2=CHCH2Br Allyl Bromide 8 PhBr Bromobenzene. They found that the S 2 rate constants N catalyzed by various crown ethers decreased in the order: oxydimethylene-bis-benzo-15-crown-5 dicyclohexyl-18-crown-6 18-crown-6. The lH NMR spectrum of the reaction mixture of phenol with benzyl bromide in toluene in the presence of. The compound is a reagent for introducing benzyl groups. S N2 backside attack is easy on the methyl group. An sp 3-hybridized electrophile must have a leaving group (X) in order for the reaction to take place. Test each of the following organic bromides similarly: 2-bromobutane, t-butyl bromide, allyl bromide (3-bromopropene), benzyl bromide, and bromobenzene. PROBLEM 6-24: 3-Bromocyclohexene is a secondary halide, and benzyl bromide is a p 6. benzyl chloride. SN2 reaction of the amine with an excess of CH3I in the first step yields an. Adding a hydroxyl group to the double bond of butene results in an nucleophilic compound with the negative charge on the terminal carbon, this terminal carbon compound under the right conditions perhaps can participate in a substitution reaction replacing bromine (NOT sn2 sn1 pathway). Test each of the following organic bromides similarly: 2-bromobutane, t-butyl bromide, allyl bromide (3-bromopropene), benzyl bromide, and bromobenzene. EDIT: Never mind, I just realized the cyanide anions complex with zinc ions to form tetracyanozincates. " problems retrosynthesis sets retrosynthetic analysis rules of thumb Self-directed Learning sigma framework SmartWork Smartwork5 SN1/SN2 SN1/SN2/E1/E2 Solubility Solved Problems solvents Spectra Spectroscopy. A) B) C) D) E) 4. The structure of allyl bromide is shown here. So we form benzyl alcohol for this reaction. Benzyl bromide is a primary alkyl halide. Interactive 3D animation of SN2 substitution at a benzylic centre for students studying advanced school chemistry and University chemistry. The molecule consists of a benzene ring substituted with a bromomethyl group. 52) Any involvement of solvent in the reaction cannot be detected in the rate law because the con-centration of the solvent cannot be changed. Search Search. S N 1: The S N 1 mechanism involves a carbocation intermediate, and both allylic and benzylic carbocations have resonance, which increases the stability of their carbocations, and speeds up the rate of S N 1 reaction. independent. Summary of Sn1 and Sn2 reactions and the types of molecules and solvents that favor each. Bromobenzene doesn’t react under either of these conditions. Bromobenzene is an aryl halide. Nucleophilic tendencies of thiophenoxide, azide, chloride, and ethoxide ions toward a series of alkyl and benzyl bromides, as well as substituent effects on rates of SN2 reactions of benzyl. The method of conductivity is used to measure the degree of dissociation of alkali halide in acetone. Concurrent SN1 and SN2 reactions in the benzylation of pyridines quinuclidine ring with benzyl bromide and at the quinoline ring with benzhydrylium ions (diarylcarbenium ions). 1-chlorobutane 3-bromopentane 3-bromocyclohexene benzyl bromide 2-methyl-3-bromopropene 1-bromo-2,2-dimethylpropane chlorobenzene bromocyclopentane 1-bromo-3-methyl-2-butene undergo an SN1 or SN2 mechanism or possibly both. 32 videos Play all Aromatic compounds | Organic Chemistry | Khan Academy Khan Academy Benzene Side Chain Reactions - Duration: 10:39. 塩化ベンジル（えんか—、benzyl chloride）は、有機合成で用いられる芳香族化合物の一種。示性式は C6H5CH2Cl、トルエンのメチル基の水素をひとつ塩素に置き換えた構造を持ち、α-クロロトルエン、クロロメチルベンゼンと呼ぶこともできる。. Benzyl bromide when kept in acetone and H 2. R in Acetic Acid 33%; HydroBromic Acid 48%; Methyl Bromo Acetate; Propionyl Bromide; Organic Bromide. CH3CH2CH2CH2ONa + NaOCH3c. Question 6 Give reasons for the following: i) Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms isocyanide as the chief products. Tert- Butyl Chloride undergoes a solvolysis reaction when it is dissolved in a polar and protic solvent. Whether an alkyl halide will undergo an S N 1 or an S N 2 reaction depends upon a number of factors. 52 Reaction Rates 1. p-Methoxybenzyl bromide was syn-thesized by the bromination14 of p-methoxybenzyl alcohol with hydrobromic acid. The major hazards encountered in the use and handling of benzyl bromide stem from both its toxicologic and physical/chemical properties. Benzyl chloride, C 6 H 5 CH 2 Cl, readily undergoes an S N 1 reaction, yet it is a primary substrate. University of Illinois UIC at Chicago CHEM 232, Spring 2010 Slide Lecture 26: April 15 Allylic SN2 Faster: Two Arguments 2. Journal of Physical Organic Chemistry 1994 , 7 (5) , 234-243. 1-Bromoadamantane is a tertiary halide, yet it is 10,000 times slower than tert- butyl bromide when reacting with silver nitrate in ethanol. Scribd is the world's largest social reading and publishing site. In certain embodiments, R SN1 is unsubstituted alkyl (e. C6H5Br - Neither Sn1 or Sn2, aromatic halide. 52) Any involvement of solvent in the reaction cannot be detected in the rate law because the con-centration of the solvent cannot be changed. pdf), Text File (. Hydrobromic acid is an important chemical in chemistry, especially in inorganic chemistry. eg: The lightest benzylic carbocation 1 is called the benzyl carbocation. MARIJUANA CHEMISTRY Genetics, Processing & Potency by Michael Starks sal Introduction to Second Edition Reviewing the work on the botany and chemistry of marijuana which has appeared in the twelve years since the publication of the first edition of this work is both gratifying and dismaying. The solution temperature (0o C) greatly slows Sn1 carbocation formation. It is known as an S N 1 reaction. txt) or read online for free. Reaction Heating Time 1 Table 1: S2 Reactivity Structure Name CH3CH2CH2CH2CI n-Butyl Chloride CH3CH2CH2CH2Br n-Butyl Bromide 3 CH3CHBYCH2CH3 sec-Butyl Bromide 4 (CH3)2CHCH2BI Isobutyl Bromide 5 (CH3),CCI tert-Butyl Chloride 6 CHCHCI Benzyl Chloride 7 CH2=CHCH2BT Allyl Bromide 8 PhBr Bromobenzene 7. Iodide is a good nucleophile, and if it displaces bromide or chloride, the NaBr or NaCl produced will precipitate.
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2020-08-04 22:39:18
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https://stats.stackexchange.com/questions/473681/proportionality-of-discipline-incidents-between-student-groups/473724#473724
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# Proportionality of discipline incidents between student groups
My goal is to see if one student group has a disproportionate number of discipline incidents relative to the another group/s, to what degree, and with what margin of error. I will be speaking mostly in lay terms, and my intended audience is a lay audience.
I am going to simplify the problem as such: I have two groups (not samples, complete populations) of students, M and F, for a given school.
There are 500 M students and 200 F students.
Of the M students, there are 100 with >= 1 discipline incident. Of the F students, there are 20 with >= 1 discipline incident.
So: 500/700 (~71.4%) of students are M students and 200/700 (~28.6%) of students are M students
And: 100/120 (83.333%) of total students having >=1 incident are M students, while 20/120 (16.666%) of total students having >=1 incident are F students
Are the following calculations then valid?
Ratio of M students with >=1 incident per 100 students = 83.333% / 71.4% * 100 = ~117
Ratio of F students with >=1 incident per 100 students = 16.666% / 28.6% * 100 = ~58
If this is valid, with what margin of error, given the total students in each group and the number of students with incidents in each group, can I say that this accurately represents the proportionality of incidents between these groups?
• Out of 100 people 117 have >= 1 incident? I'm not sure how to make sense out of it. (It sounds like "12 out of 10 dentists recommend Colgate toothpaste.") If you just want see which groups seem to have more incident occurrences, why not just 100 / 500 for M and 20 / 200 for F? Jun 23 '20 at 20:51
• Good question, and I understand why this was confusing. I was attempting to incorporate the size of the populations into the analysis rather than just the proportions. Jun 24 '20 at 21:16
## 1 Answer
For M students, you have 100 out of 500 of with disciplinary incidents. So the estimated proportion of M students with incidents is $$\hat p_M = 1/5$$ Similarly, for F students you have 20 out of 200 with incidents for an estimated proportion $$\hat p_F = 1/10.$$
So the two estimated proportions are $$0.2 = 20\%$$ for M and and $$0.1 = 10\%$$ for F. The sample proportions are different, but the difference is not necessarily statistically significant. You can't just look at percentages without taking sample sizes into account.
For example if you had 1 out of 5 M students with incidents and 1 out of 10 F students with incidents, that might be an interesting story, but it would be an anecdote rather than serious evidence. (Suppose a box has 5 blue marbles and 10 red ones; if I take out two at random, it would hardly be surprising to get one marble of each color.)
A proper statistical test will let you test the null hypothesis whether the rates $$p_M$$ and $$p_F$$ are the same against the alternative hypothesis that they are not equal. In R statistical software the procedure for conducting this test is called prop.test, illustrated below:
prop.test(c(100,20), c(500,200), cor=F)
2-sample test for equality of proportions
data: c(100, 20) out of c(500, 200)
X-squared = 10.057, df = 1, p-value = 0.001517
alternative hypothesis: two.sided
95 percent confidence interval:
0.04561321 0.15438679
sample estimates:
prop 1 prop 2
0.2 0.1
The P-value 0.001517 means that of the proportions for M and F were equal, then the probability of a difference greater than $$|p_M = P_F| = 0.1$$ or greater would happen with probability $$0.001517$$ (that's less than 1 chance in 600),
Thus, if the null hypothesis is true, a very rare event has happened. This description of the data causes us not to believe that the null hypothesis is true. So we say that the difference in observed rates is statistically significant.
The printout from the test also gives a 95% confidence interval for the population difference $$|p_M-p_F|$$ That confidence interval is $$(0.046, 0.154).$$ or $$0.1 \pm 0.054.$$ The number $$0.054$$ is called the margin of error of the estimated difference $$0.1.$$
Note: The argument cor=F in prop.test is to turn off 'continuity correction', which is not necessary with the number of students you have.
• This is excellent thanks! I am running the real number now in R. Jun 24 '20 at 22:08
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2022-01-22 21:56:32
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https://quantumcomputing.stackexchange.com/questions/14525/why-and-how-is-quantum-noise-predictable
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Why and how is quantum noise predictable?
I have been learning about quantum error correction using the zero-noise extrapolation method from this paper and have been pleased with the results. This method takes advantage of the fact that the quantum gate noise is predictable as seen, for example, in the plot below. Here, I have taken a simple 2-qubit circuit given by qc.x(0), qc.h(0), qc.cx(0,1), qc.x(0) and added noise to it in the form of pairs of CNOT gates (a noisey identity matrix). I am plotting the $$|01\rangle$$ solution for this circuit as a function of the number of CNOT pairs that I used. This is done using the qiskit noise model with the IBMQ-montreal gate error rates. Note that the real solution is 0.5.
Clearly, the noise follows a very predictable relationship which to me is unexpected. Instead, what I expected was the noise to be scattered evenly across the real solution (0.5) with the amount of scatter being proportional to the number of gates, as shown in the made up plot below.
My question is, why does the noise follow the predictable curve instead of a random scattering about the real solution? And a follow up question, for an arbitrary circuit, how can I predict what the curve will look like?
Update: I ran my noisy circuits on the real IBMQ-athens machine to see if it is anywhere close to the simulated results. The plot is below is meant to be the same as the first plot (except the maximum depth is a less because I could not exceed 900).
Clearly the noise is no where near as organized as in the simulated version. What I don't know is if that was a failure on my part to simulate the noise correctly, or if it is a problem of the qiskit noise simulator itself.
• Is the first plot coming from actual hardware or is it from using a noise model? – KAJ226 Nov 7 '20 at 23:47
• The first plot is from the qiskit noise model where I used the function NoiseModel.from_backend. – thespaceman Nov 8 '20 at 0:33
• I think that although the noise model does try to mimic the behavior of the hardware, it is still kinda fixed... what I mean by that is that on real hardware, the noise fluctuates from run to run somewhat. The gate fidelity changes from run to run rather than stay fixed like it would have with the noise model. – KAJ226 Nov 8 '20 at 0:37
• Follow up question, why is the noise model so different compared with the noise from a real QC (plot 1 with my plot 3 respectively). – thespaceman Nov 9 '20 at 18:05
• The noise on the hardware fluctuates randomly within certain range from run to run. The noise model is only meant to represent the hardware noise model in a simple sense. Now, what you can do is to do more shots... This will smoothen out the curve you have more. IBM has max number of shots of 8192 set on their device to make sure people don't take up all the time, but you can submit many jobs and average out the results. – KAJ226 Nov 9 '20 at 19:22
1 Answer
The curve you are observing seems intuitive to me, since it shows how decoherence works. As you add more gates the state of the qubits tend to the ground state asymptotically .
You can probably use T1 and T2 to predict the curve.
• I'm far from being an expert, so converting this answer in a community wiki to allow others to chip in. – luciano Nov 8 '20 at 0:04
• Interesting, so the asymptotic relationship is given by decoherence and not gate noise. Would I be correct in saying that if I isolate the gate noise, that it should look something like my second plot? – thespaceman Nov 8 '20 at 0:39
• Just a note, if you can edit your response to answer my follow up question then I can accept yours as the official answer. Thanks! – thespaceman Nov 10 '20 at 18:26
• well.. I'm not fully sure what you mean with "gate noise" in this context. Hopefully, somebody else more knowledgeable might jump in. – luciano Nov 10 '20 at 21:46
• Okay, understandable and I appreciate your help. The noise I am referring to is the noise due only to gate failure without accounting for decoherence in anyway. – thespaceman Nov 10 '20 at 23:16
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2021-05-08 06:18:16
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https://www.mysciencework.com/publication/show/light-scalar-mesons-comments-behavior-1-nc-expansion-near-nc-3-versus-nc-infinity-limit-8874eea3
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# Light scalar mesons: comments on their behavior in the 1/Nc expansion near Nc=3 versus the Nc --> infinity limit
Authors
Type
Published Article
Publication Date
May 28, 2009
Submission Date
May 28, 2009
Source
arXiv
We briefly review how light meson resonances are described within one and two-loop Unitarized Chiral Perturbation Theory amplitudes and how, close to Nc=3, light vectors follow the Nc behavior of $q\bar q$ mesons whereas light scalars do not. This supports the hypothesis that the lightest scalar is not predominantly a $q\bar q$ meson, although a subdominant $q\bar q$ component is suggested around 1 GeV at somewhat larger Nc. In contrast, when Nc is very far from 3, like in the Nc --> infinity limit, we explain again in detail why unitarization is not, a priori, reliable nor robust and why this limit should not be used to drag any conclusions about the dominant nature of physical light scalar mesons.
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2018-07-16 16:37:10
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https://collegephysicsanswers.com/openstax-solutions/cyclotron-accelerates-charged-particles-shown-figure-2270-using-results-previous
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Question
A cyclotron accelerates charged particles as shown in Figure 22.70. Using the results of the previous problem, calculate the frequency of the accelerating voltage needed for a proton in a 1.20-T field.
Question Image
$18.3 \textrm{ MHz}$
Solution Video
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2019-02-17 01:11:01
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https://terrytao.wordpress.com/tag/arzela-ascoli-theorem/
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You are currently browsing the tag archive for the ‘Arzela-Ascoli theorem’ tag.
One of the most useful concepts for analysis that arise from topology and metric spaces is the concept of compactness; recall that a space ${X}$ is compact if every open cover of ${X}$ has a finite subcover, or equivalently if any collection of closed sets with the finite intersection property (i.e. every finite subcollection of these sets has non-empty intersection) has non-empty intersection. In these notes, we explore how compactness interacts with other key topological concepts: the Hausdorff property, bases and sub-bases, product spaces, and equicontinuity, in particular establishing the useful Tychonoff and Arzelá-Ascoli theorems that give criteria for compactness (or precompactness).
Exercise 1 (Basic properties of compact sets)
• Show that any finite set is compact.
• Show that any finite union of compact subsets of a topological space is still compact.
• Show that any image of a compact space under a continuous map is still compact.
Show that these three statements continue to hold if “compact” is replaced by “sequentially compact”.
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2017-08-22 11:07:05
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{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.47651994228363037, "perplexity": 1184.0792980769018}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886110578.17/warc/CC-MAIN-20170822104509-20170822124509-00485.warc.gz"}
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https://itectec.com/ubuntu/ubuntu-resolving-dependencies-related-to-32-bit-libraries-on-64-bit/
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# Ubuntu – Resolving dependencies related to 32 bit libraries on 64 bit
I am struggling to make a 32bit application (related to adobe AIR)
work on my 64bit Ubuntu system. The application starts fine but there
is no sound. After running the application from the terminal, I've
noticed the following error messages:
/usr/lib/gtk-2.0/2.10.0/menuproxies/libappmenu.so: wrong ELF class: ELFCLASS64
Gtk-WARNING **: Failed to load type module:
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2021-11-28 21:29:31
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{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1948835551738739, "perplexity": 10219.34236077519}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358591.95/warc/CC-MAIN-20211128194436-20211128224436-00404.warc.gz"}
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https://wiki.haowen-xu.com/Mathematics/Optimization/Convex_Optimization/
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# Convex Optimization
## Cones and generalized inequalities
### Affine set, convex set and cone
• A set $$C$$ is affine if $$\forall x_1, x_2 \in C$$ and $$\forall \theta$$, we have $$\theta x_1 + (1-\theta) x_2 \in C$$.
• subspace: If $$C$$ is an affine set and $$x_0 \in C$$, then $$V = C - x_0 = \{x-x_0\,|x \in C\}$$ is a subspace.
• $$\mathbf{aff}\,C = \{\theta_1 x_1 + \dots + \theta_k x_k\,|\,x_i \in C, \sum_i \theta_i = 1\}$$, the affine hull of $$C$$.
• A set $$C$$ is convex if $$\forall x_1, x_2 \in C$$ and $$\forall \theta$$ satisfying $$0 \leq \theta \leq 1$$, we have $$\theta x_1 + (1 - \theta) x_2 \in C$$.
• $$\mathbf{conv}\,C = \{\theta_1 x_1 + \dots + \theta_k x_k\,|\,x_i \in C, \theta_i \geq 0, \sum_i \theta_i = 1\}$$, the convex hull of $$C$$.
• A set $$C$$ is a cone if $$\forall x \in C$$ and $$\forall \theta \geq 0$$, we have $$\theta x \in C$$.
• convex cone $$C$$: $$\forall x_1, x_2 \in C$$ and $$\forall \theta_1, \theta_2 \geq 0$$, we have $$\theta_1 x_1 + \theta_2 x_2 \in C$$.
• proper cone $$K$$:
• $$K$$ is convex.
• $$K$$ is closed.
• $$K$$ is solid, that is, having nonempty interior.
• $$K$$ is pointed, that is, $$x \in K, -x \in K \implies x = 0$$.
• conic hull of $$C$$: $$\{\theta_1 x_1 + \dots + \theta_k x_k\,|\,x_i \in C, \theta_i \geq 0\}$$
Some examples of proper cones:
• $$\mathbf{S}^n$$, the set of all positive semidefinite matrix, is a convex cone.
### Separating and supporting hyperplanes
• separating hyperplane: For nonempty convex sets $$C$$ and $$D$$ satisfying $$C \cap D=\varnothing$$, there exist $$a \neq 0$$ and $$b$$ such that $$a^T x \leq b, \, \forall x \in C$$ and $$a^T y \geq b, \, \forall y \in D$$. The hyperplane $$\{x\,|\,a^T x=b\}$$ is a separating hyperplane for $$C$$ and $$D$$.
• supporting hyperplane: For $$C \subseteq \mathbb{R}^n$$ and $$x_0 \in \mathbf{bd}\,C = \mathbf{cl}\,C \setminus \mathbf{int}\,C$$, if $$a \neq 0$$ satisfies $$a^T x \leq a^T x_0, \, \forall x \in C$$, then the hyperplane $$\{x\,|\,a^T x = a^T x_0\}$$ is a supporting hyperplane to $$C$$ at $$x_0$$. For any nonempty convex set $$C$$ and any $$x_0 \in \mathbf{bd}\,C$$, there exists a supporting hyperplane to $$C$$ at $$x_0$$.
• strict supporting hyperplane: if the hyperplane is a supporting hyperplane to $$S$$ at $$x_0$$, and intersects $$S$$ only at $$x_0$$.
• If a set is closed, has nonempty interior, and has a supporting hyperplane at every point in its boundary, then it is convex.
### Dual cone
The dual cone of a cone $$K$$ is defined as: $K^* = \{y\,|\,x^T y \geq 0, \, \forall x \in K\}$ Geometrically, $$y \in K^* \Longleftrightarrow -y$$ is the normal of a hyperplane that supports K at the origin.
Dual cone can be also defined for an arbitrary set $$S$$. For example:
• The dual cone of subspace $$V \subseteq \mathbb{R}^n$$ is its orthogonal complement $$V^\perp=\{y\,|\,v^Ty=0, \, \forall v \in V\}$$
The properties of dual cones (given a cone $$K$$):
• $$K^*$$ is closed and convex.
• $$K_1 \subseteq K_2 \implies K_2^* \subseteq K_1^*$$.
• $$K$$ has nonempty interior $$\implies K^*$$ is pointed.
• $$\mathbf{cl}\,K$$ is pointed $$\implies K^*$$ as nonempty interior.
• $$K^{**} = \mathbf{cl}\left(\mathbf{conv}\,K\right)$$. This implies if $$K$$ is a proper cone, then $$K^{**} = K$$.
where $$\mathbf{cl}\,K$$ is the closure of $$K$$, and $$\mathbf{conv}\,K$$ is the convex hull of $$K$$.
### Generalized inequality
Generalized inequality is a partial ordering defined by a proper cone $$K$$: \begin{align} x \preceq_K y &\Longleftrightarrow y - x \in K \\ x \prec_K y &\Longleftrightarrow y - x \in \mathbf{int}\,K \end{align}
where $$\mathbf{int}\,K$$ means the interior of $$K$$. When $$K=\mathbb{R}^n_+$$ ($$\mathbb{R}_+$$ is the non-negative half of $$\mathbb{R}$$), the induced $$\prec_K$$ and $$\preceq_K$$ is the ordinary elementwise inquality in $$\mathbb{R}^n$$.
$$\preceq_K$$$$\prec_K$$
if $$x \prec_K y$$, then $$x \preceq _Ky$$
additiveif $$x \preceq_K y$$ and $$u \preceq_K v$$, then $$x+u \preceq_K y+v$$if $$x \prec_K y$$ and $$u \preceq_K v$$, then $$x+u \prec_K y+v$$
transitiveif $$x\preceq_K y$$ and $$y \preceq _Kz$$, then $$x \preceq_K z$$if $$x\prec_K y$$ and $$y \prec _Kz$$, then $$x \prec_K z$$
reflective$$x \preceq_K x$$$$x \not\prec_K x$$
antisymmetricif $$x \preceq_K y$$ and $$y \preceq_K x$$, then $$x = y$$
preserved under limitsif $$x_i \preceq_K y_i$$ for $$i = 1, 2, \dots$$, $$x_i \to x$$ and $$y_i \to y$$ as $$i \to \infty$$, then $$x \preceq_K y$$
if $$x \prec_K y$$, then $$\exists u, v$$ small enough, s.t. $$x+u \prec_K y+v$$
### Dual generalized inequalities
If $$K$$ is a proper cone, then $$K^*$$ is also proper. The generalized inequality $$\preceq_{K^*}$$ is refered as the dual of the generalized inequality $$\preceq_K$$. We further have:
• $$x \preceq_K y \Longleftrightarrow \lambda^T x \leq \lambda^T y, \, \forall \lambda \succeq_{K^*} 0$$
• $$x \prec_K y \Longleftrightarrow \lambda^T x < \lambda^T y, \, \forall \lambda \succeq_{K^*} 0, \lambda \neq 0$$
### Minimum and minimal elements
• $$x \in S$$ is the minimum element of $$S$$ (w.r.t. $$\preceq_K$$) $$\Longleftrightarrow$$ any one of:
• for every $$y \in S$$, we have $$x \preceq_K y$$
• $$S \subseteq x + K$$
• $$x \in S$$ is a minimal element of $$S$$ (w.r.t. $$\preceq_K$$) $$\Longleftrightarrow$$ any one of:
• $$y \in S$$, $$y \preceq_K x$$ $$\implies$$ $$y=x$$
• $$(x-K)\cap S = \{x\}$$
### Minimum and minimal elements via dual inequalities
• $$x \in S$$ is the minimum element of $$S$$ (w.r.t. $$\preceq_K$$) $$\Longleftrightarrow$$ any one of:
• $$\forall \lambda \succ_{K^*} 0$$, $$x$$ is the unique minimizer of $$\lambda^T z,\,\forall z \in S$$.
• $$\forall \lambda \succ_{K^*} 0$$, the hyperplane $$\{y\,|\,\lambda^T(y-x)=0\}$$ is a strict supporting hyperplane to $$S$$ at $$x$$.
• The necessary and sufficient condition for $$x$$ to be minimal in $$S$$ (w.r.t $$\preceq_K$$):
• For a particular $$\lambda \succ_{K^*} 0$$, if $$x$$ minimizes $$\lambda^T z, \, \forall z \in S \implies x$$ is minimal.
• If $$S$$ is convex, then $$\forall$$ minimal element $$x$$, there $$\exists \lambda \succeq_{K^*}, \, \lambda \neq 0$$, s.t. $$x$$ minimizes $$\lambda^T z, \, \forall z \in S$$.
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2020-01-23 16:44:49
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http://math.stackexchange.com/questions/151313/smallest-ideal-containing-i-j
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# Smallest ideal containing $I$ & $J$
I'm trying to think of an example of a ring $A$ and ideals $I$,$J$ s.t. $I \cup J$ is not an ideal.
And what is the smallest ideal containing $I$ & $J$?
Will $A \mathbb{Z}$, $I = 2\mathbb{Z}$, and $J = 4\mathbb{Z}$ work? and the smallest ideal containing them be $\mathbb{Z}$, the integers?
Can someone add some explanation as to why this works? Thanks.
-
$2\mathbb{Z} \cup 4\mathbb{Z}=2\mathbb{Z}$ – Chris Eagle May 29 '12 at 21:21
$2\mathbf Z$ and $3\mathbf Z$ would work, though. [Check whether the union is closed under addition.] – Dylan Moreland May 29 '12 at 21:24
Ironically, most choices of $I$ and $J$ would work with $A = \mathbb{Z}$; you had the misfortune of making one of the special choices that doesn't. – Hurkyl May 29 '12 at 21:26
Hint 1: Try $2\mathbb{Z}\cup 3\mathbb{Z}$ in $\mathbb{Z}$.
Hint 2: Ideals are additively closed, so at the very least the smallest ideal containing $I$ and $J$ also has to contain $I+J$...
You could try the ideals $I=3\mathbb{Z}$ and $J=5\mathbb{Z}$. Then their union is not an ideal since it is not closed under addition. You have $3,5\in I\cup J$, but $8\notin I\cup J$, for example.
Secondly, the smallest ideal containing $I$ and $J$ is $I+J$. A brief proof can be found here as problem 2a.
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2015-04-21 15:33:43
|
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http://thox.agenziafunebretassoni.it/congruent-symbol-on-mac-keyboard.html
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### Congruent Symbol On Mac Keyboard
Program information. If your keyboard doesn't have a dedicated Numpad you may have to use a Fn key in combination with a Num Lock key. What you see when you first glance at the iPhone or iPad keyboard are the regular letters, numbers, and symbols you're most likely to use in everyday communication. 42 released 12/11/2018: Fixed: Certain symbols incorrect if saved on Mac and loaded on Windows or vice versa. Luckily, it’s easy to show all Mac key symbols at once: Go to System Preferences ➙ Keyboard. ⌥ N and ⇧ A. gradians / gons. To insert one, click on it, then click Insert. Check the box next to "Show keyboard and emoji viewers in menu bar". You can just copy and paste the different approximately symbols below. Note: you could also use keyboard shortcut to. Arithmetic & Algebra. To open the menu, click the Insert tab in the Ribbon, then click Symbol: You'll see the Symbol menu: From here, you can scroll through hundreds of symbols. Open System Preferences-> Keyboard. Here, you'll see all kinds of categories on the left: Emoji, Arrows, Currency Symbols, etc. Method 1 Unicode value: 8776 Method 2 If you're using HTML, you can just type in (≈) and it will display the character. However, if you plan to open a document on multiple platforms, you may want to use one of Adobe's Euro font. There's a whole lot more, however, lurking just beneath the surface, including accented (diacritic) characters, ligatures, extended punctuation, and special symbols. There are many keyboard techniques that will help you use Sketchpad productively. You can help protect yourself from scammers by verifying that the contact is a Microsoft Agent or Microsoft Employee and that the phone number is an official Microsoft global customer service number. Sometimes students or those who deal with mathematics, physics or various kinds of calculations may need to type a degree sign, but we do not have one directly on our keyboard. OS X has, however, the power to input well in excess of the characters residing on those eighty, or so, individual keys through a handy system of virtual input. Use the keyboard shortcut - ALT + 30 (you need to hold the ALT key and then press 30 from the numeric keypad of your keyboard). If you don’t have this font, and even if you do, I recommend downloading and installing the free Stix fonts which have lots, and lots of mathematical characters. To insert the European Community currency symbol, you must use an operating system that includes the symbol in its fonts—such as Windows 98 or later, or Mac OS 8. Linux is different and to achieve the same, hold down Ctrl+Shift and type u+00b0, though if you are writing for html, the correct. If you don't already have one, you can create one right now. NOTE: If you have the International keyboard activated, you will only be able to input codes with the ALT key on the left side of the keyboard. If you want to know number of some Unicode symbol, you may found it in a table. Apple may provide or recommend responses as a possible solution based on the information provided; every potential issue may involve several factors not detailed in the conversations captured in an electronic forum and Apple can therefore provide no guarantee as to the. How to Install Rupee Font Download the Rupee Font from Rupee_Foradian. The closely related code point U+2262 ≢ NOT IDENTICAL TO (HTML ≢) is the same symbol with a slash through it, indicating the negation of its mathematical. There are two great ways to write exponents on a Mac! First, you can use the classic caret symbol (above the 6) by pressing the shift key and 6. They are named as Option with the small name "alt" on it. Make sure that the Key Repeat slider isn't set to Off. rightwards arrow. In addition to other Mac keyboard shortcuts, you can also use option key in combination with other keys to insert special symbols like trademark, copyright and currencies. Multiplication sign. This operator represents the multiplication of two quantities, e. By default the table is sorted with the first column "Alt + Key". This superscript. o : ô ö ò ó œ ø ō õ. Example: Cyrillic capital letter Э has number U+042D (042D - it is hexadecimal number), code ъ. Try a different character or try to insert it in a different application. On the Insert tab, in the Symbols group, click the arrow under Equation, and then click Insert New Equation. Have not found euro symbol on yahoo, the sub numbers as used in H2O, congruent symbol. Make sure that the Key Repeat slider isn't set to Off. After signing in or signing up for a Parallels Desktop account, Windows 10 will pop up in a new window. 04, in system setting -> keyboard configuration, choose options, there is a option called 3rd level (in spanish Tecla para seleccionar el 3er nivel), I chose the numeric intro that is the right key beside the right mac cmd key, and there you have it presing that key plus 2 I get the at symbol. You can insert special characters in your documents and presentations without having to remember all those Alt-codes by using Google Docs and Slides easy-to-use character insertion tool. I also reached out to retouchers in the industry to get their take on where apple is going. What is an asterisk used for on a computer? Below are examples of how the asterisk is used on a computer. Contains both 1 - 256 and 0128 - 0256 code ranges. To use this symbol press down shift and the number 6 on your keyboard. radians angle unit. And they would likely haemorrhage market share, revenue and profit because selling 100,000 $10,000 phones doesn't net you nearly as much cash as selling 29 million$1,000 phones. Tech support scams are an industry-wide issue where scammers trick you into paying for unnecessary technical support services. And you can type it right from your keyboard. If you open the Character Map and select the Registered Trademark symbol, you'll see that you could enter the ® symbol by pressing Alt+0174. The closely related code point U+2262 ≢ NOT IDENTICAL TO (HTML ≢) is the same symbol with a slash through it, indicating the negation of its mathematical. Synonyms for congruent in Free Thesaurus. Algebra symbols Algebra symbols. active oldest votes. Keyboard shortcuts. After signing in or signing up for a Parallels Desktop account, Windows 10 will pop up in a new window. I have been using the mac for a long time, earlier I used Entourage as my email client, then came the Outlook 2011 for Mac and now its Outlook 2016 for Mac. We have Keyboard Tester to check all the keyboard keys online. right from your keyboard. Try Keyboard Checker below, which is to test and helps to fix broken keys. In the Microsoft Windows operating system, the angle sign is treated as a symbol. 4 Calculus, Vector Calculus, Differential Equations. To choose the appropriate Math function symbol: a. In List view, collapse a folder. gradians / gons. They are organized into seven classes based on their role in a mathematical expression. Click on the Insert Symbols (Math) tools. Image Cool Illustration Eye Catchy 2016 Image Minimal Logo Elegant Icon Logo Corporate Unique Logo Template Branding. Disclosure: This post may contain affiliate links, meaning, at no additional cost to. Note: If a +1 button is dark blue, you have already +1'd it. Buy a comprehensive geometric formulas ebook. The tilde is used in some languages as a diacritical mark. How to use the Indian Rupee Font in MS Application. In Word, you can insert mathematical symbols into equations or text by using the equation tools. Both systems offer all the major. S and Window O. I can't get that equal part of. AL Go! Generates a HelloWorld project. ) The American Standard Code for Information Interchange, or ASCII code, was created in 1963 by the "American Standards Association" Committee or "ASA", the agency changed its name in 1969 by "American National Standards Institute" or "ANSI" as it is known since. The * is the same as the x multiplication sign, but it is often used in computer terminology as a result of possible confusion with the letter 'x'. Just copy the font and paste it in "Fonts" folder in the Control Panel) Done. The closely related code point U+2262 ≢ NOT IDENTICAL TO (HTML ≢) is the same symbol with a slash through it, indicating the negation of its mathematical. But to type it, you need to have a font that contains it. 2 Binary operators. Definition of corresponding in the Idioms Dictionary. Uncheck Open Bluetooth Setup Assistant at startup if no mouse or trackpad is detected. Select all occurrences. To see all Mac key symbols, you need to select Show Emoji & Symbols option from the same language flag menu, or use a shortcut Control + Cmd + Space. Linux is different and to achieve the same, hold down Ctrl+Shift and type u+00b0, though if you are writing for html, the correct. It is exasperating to me that if I need to tweak it a bit, I move a polyline vertex between two arcs and it distorts the shape of the attached arcs so that I t. Use unicode Degree symbols in a html document or copy paste the character. Use the keyboard shortcut - ALT + 30 (you need to hold the ALT key and then press 30 from the numeric keypad of your keyboard). mathl-alt 4. On iPhone and iPad. In ubuntu 12. The "does not equal" or "is not equal to" sign is an expression of the inequality between two different numbers, variables, integers, or concepts. The word pairs were presented for 5 s or until participants responded. gradians / gons. After Windows 10 has installed on your Mac, you'll be directed to sign into your Parallels Desktop account. Hold Command-V to paste the character repeatedly. Letters are. The shortcuts Help topic is available when you're working on a computer, but not in Mastering's Mobile view on mobile devices. π Ratio of circumference to diameter of a circle. You could use \mathbb {Z} to represent the Set of Integers! improve this answer. Keyboard shortcuts. If you want to fix this issue, then you are in the right place. You can check every keyboard buttons on your. Statistical symbols. Multiplication, multiply. There are a lot of special characters and symbols which are currently supported. AL Go! Generates a HelloWorld project. Keyboard Shortcut. Page Content Greek Letters New Page Common Arithmetic & Algebra Superscript and Subcript Statistic…. ⬈ ⬉ ⬊ ⬋ ⬌ ⬍ ⏎ ⬎ ⬏ ⬐ ⬑ ☇ ☈ ⍃ ⍄ ⍇ ⍈ ⍐ ⍗ ⍌ ⍓ ⍍ ⍔ ⍏ ⍖ ⍅ ⍆. [54%off]Geometric Eagle Illustration. I've been using the Kanex keyboard for a week now and have found it to be excellent. Apple may provide or recommend responses as a possible solution based on the information provided; every potential issue may involve several factors not detailed in the conversations captured in an electronic forum and Apple can therefore provide no guarantee as to the. A checkmark (also known as checked, tick or checkbox in English) is a mark ( , , , √) used to indicate the concept of "yes", to indicate that "yes, it has been verified" and also, "yes, that is the correct answer". That is, the Alt key and the 'x' key at the same time. You can also use the Keyboard Viewer to type special characters and symbols. Press question mark to learn the rest of the keyboard shortcuts. It's designed to work with the MacBook Pro, MacBook Air, iMac, and Apple Wireless Keyboard. The Symbol will display, select (normal text) in the box of Font and choose Mathematical Operators in the box of Subset. " Select this, and an image of a keyboard will appear on your screen. The Unicode entries for superscript are CTRL+SHIFT+U and then 207x, where X is the number you want in superscript. Multiplication sign. Free Blue UI icons of science for user interface and graphic design projects. Press question mark to learn the rest of the keyboard shortcuts. You can insert special characters in your documents and presentations without having to remember all those Alt-codes by using Google Docs and Slides easy-to-use character insertion tool. 360° = 400 grad. Middle dot (multiplication). Use the keyboard shortcut - ALT + 30 (you need to hold the ALT key and then press 30 from the numeric keypad of your keyboard). Buy Belkin YourType Bluetooth Wireless Keyboard for Mac - F8T067TT online on Amazon. The aluminum finish is first class and the black keys look better than the white keys on my old Apple keyboard. Below are the steps to insert the delta symbol (solid triangle/arrow symbol) in Excel using a keyboard shortcut: Select the cell in which you want to insert the degree symbol. The keyboard is rarely the friendliest path to symbols and special characters. If you like this Page, please click that +1 button, too. ∨ logical or = vee. The closely related code point U+2262 ≢ NOT IDENTICAL TO (HTML ≢) is the same symbol with a slash through it, indicating the negation of its mathematical. Mathematical Symbols Press/hold Alt key while typing the - Starr. Alt 0153 ™ Alt Code Symbol Alt 231 τ Alt 0184 ¸ Alt 123 { Alt 0203 Ë Alt 0235 ë Filled Arrows Alt 40 ( Alt 232 Φ Web Marketing Alt 125 } Alt 0204 Ì Alt 0236 ì Alt 16 Alt 41 ) Alt 237 φ by Editing Alt 0205 Í Alt 0237 í Alt 17 Plus or Minus Alt 234 Ω www. The exponent symbol on a keyboard is known as ^. Available in png and vector. You can change the filtering based on any column or enter the description in the search box and the results will be automatically filtered. If a key isn't designed to repeat in the app you’re using, follow these steps: Select the character that you want to copy. Wingdings Checkmark or Tick Symbol. With the MacBook Pro, it has become very clear to me that their products aren't congruent with the needs of professionals with the direction of their forced change. Latin letter C c. grads angle unit. However, if you plan to open a document on multiple platforms, you may want to use one of Adobe's Euro font. 11 synonyms for congruent: compatible, agreeing, according, consistent, identical, coinciding. Just click on the symbol to get more information such as fractions symbol unicode, download fractions emoji as a png image at different sizes, or copy fractions symbol to. → Special ASCII characters in HTML. This superscript. As you gain experience in Sketchpad, you'll naturally seek ways to maximize your efficiency and productivity within the software environment. In this list below, on the right side of colon (:) is the key you need to press on your iPhone's keyboard to see the symbols that are on the left of colon. S and Window O. You can just copy and paste the different approximately symbols below. You can change the filtering based on any column or enter the description in the search box and the results will be automatically filtered. why math symbols are used. 04, in system setting -> keyboard configuration, choose options, there is a option called 3rd level (in spanish Tecla para seleccionar el 3er nivel), I chose the numeric intro that is the right key beside the right mac cmd key, and there you have it presing that key plus 2 I get the at symbol. In the first page of Disquisitiones Arithmeticae Gauss says that he has adopted this symbol from Legendre but Legendre used this symbol to show both equality and congruence while Gauss uses this symbol only for congruence to distinguish it from equality. com Alt 28 ∟ Alt 0206 Ï Alt 0238 î. Indeed, the Special Character flyout palette in Adobe InDesign CS3 is a great shortcut to. In List view, expand a folder. Best Answer: The symbol is in Unicode. This command does just that: it enters edit mode if you've selected a symbol or exits exit mode if you're already editing a symbol. HTML Math Symbols, Math Entities and ASCII Math Character Code Reference. Finding what you want in this massive list of symbols can be quite difficult. The congruent symbol is an equals sign (=) with a tilde (~) above it. View and Download Silvercrest SBT 3. Sometimes students or those who deal with mathematics, physics or various kinds of calculations may need to type a degree sign, but we do not have one directly on our keyboard. To open the menu, click the Insert tab in the Ribbon, then click Symbol: You'll see the Symbol menu: From here, you can scroll through hundreds of symbols. Available in png and vector. Don't have an account. For more information, click the links below. 648 respectively, F(4,13) = 7. It offers a myriad of symbols, characters, symbols, languages, and more. Place your cursor where you want to type the degree. By default the table is sorted with the first column "Alt + Key". 360° = 2π rad. Here's how you can insert special characters into your documents. Keyboard Tester is an online tool to test on desktop, laptop, and MAC computer keyboard keys. , gMail, Docs), a login window opens when you click on +1. The PUA allows font developers to assign private Unicode values to glyphs that don't map to existing code points. Free Blue UI icons of science for user interface and graphic design projects. ) The American Standard Code for Information Interchange, or ASCII code, was created in 1963 by the "American Standards Association" Committee or "ASA", the agency changed its name in 1969 by "American National Standards Institute" or "ANSI" as it is known since. Congruent Symbol +. Click the arrow next to the name of the symbol set, and. A2A by Chew Zhi Heng. The triple bar, ≡, is a symbol with multiple, context-dependent meanings. AL Go! Generates a HelloWorld project. Don't have an account. Typing the trademark symbol next to symbols, designs or words indicates that you consider yourself the owner of those unique symbols, designs or words. On a Mac, you'll need to use the Character Viewer instead of using Alt codes. In the first page of Disquisitiones Arithmeticae Gauss says that he has adopted this symbol from Legendre but Legendre used this symbol to show both equality and congruence while Gauss uses this symbol only for congruence to distinguish it from equality. Use the keyboard shortcut. Most obviously, the square function can be used to find the area of squares. Pimp your Instagram and Facebook profiles, or create some useful text symbols like umlauts, copyright, trademark, registered sign, euro, pound, etc. ttf (Click Right Mouse " SAVE TARGET AS") Install to Fonts folder of your System (It is easy. Case in point, the Pi. For example, you can insert the Registered Trademark symbol ® by pressing Alt+Ctrl+R. HTML Math Symbols, Math Entities and ASCII Math Character Code Reference. The shortcuts Help topic is available when you're working on a computer, but not in Mastering's Mobile view on mobile devices. I've been using the Kanex keyboard for a week now and have found it to be excellent. To choose the appropriate Math function symbol: a. First select the symbol then you can drag&drop or just copy&paste it anywhere you like. Source(s): On a Mac, always check the help menu on the Finder page or on the pages of the different apps. I'm sure if Apple wanted, they could charge starting at \$10,000 for their iPhones, then they would start falling into the status symbol/ aspirational brand. Select Shortcuts from the answer box toolbar. You can help protect yourself from scammers by verifying that the contact is a Microsoft Agent or Microsoft Employee and that the phone number is an official Microsoft global customer service number. 360° = 2π rad. left-right arrow. I can not imagine that Android OS does not have the possibility to let the softkeyboard start in the number/symbol code and give the user the ability to switch back to letters? Two congruent triangles Is there a spell, magical item, or. This equation editor lets you enter various equations or symbols and keep them inline with your text. Choose Apple ( ) menu > System Preferences. Available in png and vector. S and Window O. 11 synonyms for congruent: compatible, agreeing, according, consistent, identical, coinciding. Your Mac's keyboard likely comes equipped with keys for all your most commonly-typed characters, such as your local alphabet, numbers and a range of common punctuation and basic symbols. " However, it is informally referred to as the "squiggly" or "twiddle. There are other keyboard. The other option is to hold the -alt- button while you press 1-3-0 on your number pad and see what comes up. If your mac is already booting then, Open System Preferences. Both systems offer all the major. Below is the complete list of Windows ALT key numeric pad codes for geometric shape symbols, their corresponding HTML entity numeric character references and, when available, their corresponding HTML entity named character references. The multiplication symbol, for one, is not accessible by a keyboard shortcut. Click on the Insert Symbols (Math) tools. Press J to jump to the feed. a with ring above. This is the currently selected item. In this list below, on the right side of colon (:) is the key you need to press on your iPhone's keyboard to see the symbols that are on the left of colon. In the Microsoft Windows operating system, the angle sign is treated as a symbol. Assign a keyboard shortcut to quickly jump Platforms: Mac. To use this symbol press down shift and the number 6 on your keyboard. It is exasperating to me that if I need to tweak it a bit, I move a polyline vertex between two arcs and it distorts the shape of the attached arcs so that I t. 0 ports, one at each end of the keyboard (like M2452 and M7803). You can check every keyboard buttons on your. While the keyboard on a Mac looks the same as every other keyboard on the face of the planet (except for the command and option buttons, I suppose), it comes packed with a plethora of hidden symbols that users can make use of. Best Answer: The symbol is in Unicode. your question appears to be about notation rather than tex so may be off topic, but I'd normally denote the set of integers by \mathbb {Z} (with N being natural numbers rather than integers. - Mobius Pizza Mar 7 '12 at 14:20. If you want to know the keyboard equivalent of a symbol, you will find that hovering over a symbol or button will show you the symbol's code. Insert all the mathematical symbols in Microsoft Word with help from a computer. The trademark symbol appears as the letters "TM" in a superscript font. To see all Mac key symbols, you need to select Show Emoji & Symbols option from the same language flag menu, or use a shortcut Control + Cmd + Space. For many of the symbols below, the symbol is usually synonymous with the corresponding concept (ultimately an arbitrary choice made as a result of the cumulative history of mathematics), but in. I have a Mac and I know about the Alt+___, like to type ≠, ≤, ≥, etc The thing is, I was with my friend and she pressed a key combo and it popped up a window with more symbols like congruency and stuff. Here is a list of common shortcuts for inputting symbols. LaTeX symbols have either names (denoted by backslash) or special characters. Click the arrow next to the name of the symbol set, and. 1 Unary operators. Keyboard Shortcut Listing Special Characters The Wolfram Language not only has systemwide support for arbitrary Unicode characters, but also includes nearly a thousand carefully designed characters for mathematical notation and technical presentation — all fully integrated into the Wolfram Language's input, output, and graphics. The other option is to hold the -alt- button while you press 1-3-0 on your number pad and see what comes up. A mathematical concept is independent of the symbol chosen to represent it. " Select this, and an image of a keyboard will appear on your screen. A with ring above. I also had no trouble "pairing" the keyboard with my Mac (the time from unpacking to typing took about five minutes). For more information, click the links below. How to Type the Trademark Symbol on a Macbook. Click the special character you want to insert into the document. To get the letter, character, sign or symbol "|": ( vertical-bar, vbar, vertical line or vertical slash ) on computers with Windows operating system: 1) Press the "Alt" key on your keyboard, and do not let go. The preferred way of symbolizing "approximately equal to" is to use the ≈ symbol, the ~ symbol, or (in some cases) the ≅ symbol (see note 2). To open the menu, click the Insert tab in the Ribbon, then click Symbol: You'll see the Symbol menu: From here, you can scroll through hundreds of symbols. Return to Math Symbols Page Go to the About the Codes section to see how they are implemented. ∨ logical or = vee. active oldest votes. 30 Keyboard Shortcuts for Microsoft Word. Where just to show you briefly here, notice here there's a less than or equal sign, greater than equal sign, these aren't symbols that I can't get from my keyboard. To insert Greek letter type Ctrl+G ( Command G on Mac OS) and then type Latin letter mentioned in the table below. There are two great ways to write exponents on a Mac! First, you can use the classic caret symbol (above the 6) by pressing the shift key and 6. - David Carlisle Mar 12 '14 at 1:01. If you're posting in a forum, it will depend if it's … Copy and Paste the Approximately Symbols from here Read More ». 0 A1 operating instructions manual online. This page is intended to supply a list of some useful symbols separated by topic so they can be found quickly without the need to search in the Unicode reference tables. ☑ Step 3: Copy and paste Math Symbols text wherever you want. Cmd-1, Cmd-2, Cmd-3, Cmd-4. A checkmark (also known as checked, tick or checkbox in English) is a mark ( , , , √) used to indicate the concept of "yes", to indicate that "yes, it has been verified" and also, "yes, that is the correct answer". Click Ok; If on the startup then nothing would work. Is there anywhere a "key& quot; to the SilverCres t SBT 3. I have a macbook pro model 2,2 I found a way to map the 3rd symbol easily. Note on Fonts. The exponent symbol on a keyboard is known as ^. Finding what you want in this massive list of symbols can be quite difficult. Just copy the font and paste it in "Fonts" folder in the Control Panel) Done. On a Mac, you'll need to use the Character Viewer instead of using Alt codes. Besides the symbols mentioned above, dozens more can be typed using a regular Apple keyboard. Apple has it's own method and doesn't rely on these codes. If the mouse is slowing you down, this complete list includes the most useful keyboard shortcuts to perform tasks on Windows 10 a little faster. Alt-Codes can be typed on Microsoft Operating Systems: First make sure that numlock is on, Then press and hold the ALT key, While keeping ALT key pressed type the code for the symbol that you want and release the ALT key. This page is intended to supply a list of some useful symbols separated by topic so they can be found quickly without the need to search in the Unicode reference tables. Double-click a symbol to "type" it, drag-and-drop it to a text field, or right-click and select Copy Character Info. Inserting all the mathematical symbols in Microsoft Word is something that you can do right from your keyboard. It adds 28 keys to your compatible Apple device, including number keys, function keys, and document navigation controls. a : à á â ä æ ã å ā. You can just copy and paste the different approximately symbols below. Second, many specific writing platforms (such as Apple's Pages and Google Docs) have the option of using a superscript, which helps you write something in the exponent's position. Legal Dictionary. Algebra symbols Algebra symbols. a with circumflex. Fast and free shipping free returns cash on delivery available on eligible purchase. To Activate the Keyboard Viewer: 1. Close • Posted by 1 minute ago. Choose Apple ( ) menu > System Preferences. Make sure that the Key Repeat slider isn't set to Off. Apple has it's own method and doesn't rely on these codes. Multiplication sign. Uncheck Open Bluetooth Setup Assistant at startup if no mouse or trackpad is detected. The mean RTs for each. Type a c within parentheses and it may automatically convert it to the copyright symbol. After Windows 10 has installed on your Mac, you'll be directed to sign into your Parallels Desktop account. After signing in or signing up for a Parallels Desktop account, Windows 10 will pop up in a new window. since I am writing blog post that hosted by Github with Editor Atom, and use plugin markdown-preview-plus and mathjax-wrapper, and use mathjax Javascript display the math symbols on the web page. 360° = 400 grad. I have a macbook pro model 2,2 I found a way to map the 3rd symbol easily. Open or Start software keyboard in number or symbol mode. Apple may provide or recommend responses as a possible solution based on the information provided; every potential issue may involve several factors not detailed in the conversations captured in an electronic forum and Apple can therefore provide no guarantee as to the. The easiest way to use the congruent symbol is to copy it from another source that uses the symbol and paste it where you need it. So here it is: In windows, apparently you hold down Alt+0176. Geometry word problem: the golden ratio. zip file, then search for the product folder or product file. Click the symbol that you want. Doing some math? It's not so hard with an HTML plus sign or minus sign. Show Keyboard Viewer. I spoke with Pratik Naik, from Solstice Retouch and he had the following to say:. What's the meaning of the Theta » Theta This page is about the meaning, origin and characteristic of the symbol, emblem, seal, sign, logo or flag: Theta. The chart also shows math symbols such as greater than or equal to and not equal to, plus percent, square root, per million, infinity, therefore, and more. Pinterest. I can not imagine that Android OS does not have the possibility to let the softkeyboard start in the number/symbol code and give the user the ability to switch back to letters? Two congruent triangles Is there a spell, magical item, or. Somewhat confusingly, the # sign is also sometimes referred to as a pound sign on a keyboard or phone, and this can be a good way to remember that you should hit Option and the pound sign on a Mac to type a British pound symbol. Latex symbols in Math mode. Tech support scams are an industry-wide issue where scammers trick you into paying for unnecessary technical support services. The Mac app is finally stable enough. The triple bar character in Unicode is code point U+2261 ≡ IDENTICAL TO (HTML ≡ · ≡). The triple bar, ≡, is a symbol with multiple, context-dependent meanings. Sponsored Links. I can't get that equal part of. (no standardized keyboard entry method). N400 was elicited by congruent and noncongruent Chinese idioms. Yahoo; Get our app. First select the symbol then you can drag&drop or just copy&paste it anywhere you like. However, I haven't ever seen the second symbol you provided. In List view, expand a folder. 360° = 2π rad. To help, Excel divides the symbols into sections, which you can. You can just copy and paste the different approximately symbols below. I have gathered different symbols you cn copy and use. Key shortcuts, Unicode and UTF-8 tables. Click Keyboard. Keyboard Shortcut. Then select the Control Click to Select Object Behind (Windows) or Command Click to Select Objects Behind (Mac OS) check box in the Selection area. How can I write a ° (degree) symbol in LaTeX? The \degree command is provided by the gensymb package, so if you add: \usepackage {gensymb} to your preamble, that should enable the command. ALT Codes for geometric shape symbols. Corresponding parts of congruent triangles are congruent;. Click the arrow next to the name of the symbol set, and. math symbols: Asymptotic ≈ Degree symbol ° Delta: Δ : Division sign ÷ Fraction 1/2: ½ : Fraction 1/4: ¼ : Fraction 3/4: ¾ : Greater than > Greater than or equal ≥ Infinity symbol ∞ Left Angle Bracket 〈 Less than < Less than or equal ≤ Micro: µ : Multiplication sign × not symbol ¬ Ohm sign: Ω : Per Mille (1/1000) sign ‰ Pi. Inserting the angle symbol outside of a program requires accessing the Character Map utility. In the first page of Disquisitiones Arithmeticae Gauss says that he has adopted this symbol from Legendre but Legendre used this symbol to show both equality and congruence while Gauss uses this symbol only for congruence to distinguish it from equality. 04, in system setting -> keyboard configuration, choose options, there is a option called 3rd level (in spanish Tecla para seleccionar el 3er nivel), I chose the numeric intro that is the right key beside the right mac cmd key, and there you have it presing that key plus 2 I get the at symbol. Use the Print Screen key on your keyboard on a PC (or Apple Command key + Shift + 3 on a Mac) to grab the picture of the triangle, crop it so that the picture fits the canvas, and save the file as triangle1 in png or jpg format. Latin letter C c. Release the ALT key. This equation editor lets you enter various equations or symbols and keep them inline with your text. Code then, Alt + x. For Windows machines, choose the appropriate symbols category and locate the division symbol. They will sell you a patch that allows it in yahoo but I refuse topay the dirty bastards ! If if need it that bad I type my message in notepad then send it as an attachment !. GitHub is home to over 40 million developers working together to host and review code, manage projects, and build software together. ttf (Click Right Mouse " SAVE TARGET AS") Install to Fonts folder of your System (It is easy. Symbols are a concise way of giving lengthy instructions related to numbers and logic. Financial Dictionary. Great use of spacing within the piece to make a second rectangle congruent to the sides of the painting. Now you can continue entering you text. Press Option Shift 8. Greek Letter Name. Entering special characters, including foreign currency symbols, fractions, and emoji, is a cumbersome task on most physical. 1 Special Typography Symbols. There are a lot of special characters and symbols which are currently supported. The multiplication symbol, for one, is not accessible by a keyboard shortcut. Accents include acute, grave, circumflex,. We have Keyboard Tester to check all the keyboard keys online. For example, H₂0 uses CTRL+SHIFT+U and then 2082. Use your numeric keypad with your NUM LOCK on and you will be good to go! plus-or-minus sign. The shortcuts Help topic is available when you're working on a computer, but not in Mastering's Mobile view on mobile devices. There are many keyboard techniques that will help you use Sketchpad productively. News and media publications have also adopted the new INR symbol. Insert all the mathematical symbols in Microsoft Word with help from a computer. Welcome to Useful Shortcuts, THE Alt Code resource!. There are a lot of special characters and symbols which are currently supported. By Mikey Campbell Monday, November 10, 2014, 05:43 am PT (08:43 am ET) After years of waiting, Mac faithful can finally purchase an ultra high. 30 Keyboard Shortcuts for Microsoft Word. radians angle unit. Select the next icon in Icon and List views. In this article, we provide the complete list alt code shortcuts for Mac to insert symbols. Congruent Graphics. Click Keyboard. In addition to other Mac keyboard shortcuts, you can also use option key in combination with other keys to insert special symbols like trademark, copyright and currencies. Feb 28, 2017, 10:30am each OS is now so feature-congruent that the old rivalry doesn't exactly hold true anymore. Open System Preferences-> Keyboard. The Apple logo is one fo the few symbols here that can be easily typed with a typical keyboard layout: ⌥⇧K There is also a Fn modifier on modern Mac keyboards. The Triple Horn of Odin, The Horn Triskelion ©Anne Mathiasz at ShutterStock. Click Alt+= again to exit from the equitation. Press Option Shift 8. NOTE: If you have the International keyboard activated, you will only be able to input codes with the ALT key on the left side of the keyboard. Corresponding parts of congruent figures are congruent;. You can insert special characters in your documents and presentations without having to remember all those Alt-codes by using Google Docs and Slides easy-to-use character insertion tool. Such that 3 to the power 4 would be written as 3^4. Click Keyboard. Fast and free shipping free returns cash on delivery available on eligible purchase. Case in point, the Pi. Where just to show you briefly here, notice here there's a less than or equal sign, greater than equal sign, these aren't symbols that I can't get from my keyboard. contains as member. Also contains a Symbol-to-ASCII converter - AddressMunger. I am not gonna to tell you how to make all these things work together. Below are the steps to insert the delta symbol (solid triangle/arrow symbol) in Excel using a keyboard shortcut: Select the cell in which you want to insert the degree symbol. Open the math panel using the menu command Insert / Math / Math Panel (alternatively, right-click on your mathematical formula), and look up your symbol in the various tables provided there. downwards arrow. The chart also shows math symbols such as greater than or equal to and not equal to, plus percent, square root, per million, infinity, therefore, and more. Sponsored Links. Or a combination of the Option/Alt key and the Shift key (⇧). In Word, you can insert mathematical symbols into equations or text by using the equation tools. With Mac OS X 10. On my mac I've got ⌘+⌃+⌥+⇧+z set up to cycle thru U. Below table shows the complete list of Alt key shortcuts with search function. And this method can also be applied to Microsoft Excel. I have had several people in my Microsoft Word training classes asking if there is a good list of word shortcuts. You can change the filtering based on any column or enter the description in the search box and the results will be automatically filtered. Math Symbols:. Insert all the mathematical symbols in Microsoft Word with help from a computer. Keyboard Shortcut. Typing special characters with a Chromebook can be done using unicode. Session uses the Signal encryption protocol and the Loki blockchain's decentralised Service Node network to provide anonymous, end-to-end encrypted messaging without relying on central servers. It is also a common coordinate degree sign. In this post, I am gonna show you how to write Mathematic symbols in markdown. Choose Apple ( ) menu > System Preferences. Switch to the numbers and symbols keyboard and press the ALT key. → Special ASCII characters in HTML. To use this symbol press down shift and the number 6 on your keyboard. GitHub is home to over 40 million developers working together to host and review code, manage projects, and build software together. Note: you could also use keyboard shortcut to. From the iOS keyboard on your iPhone or iPad: Android. Witthaya Prasongsin / Getty Images The information in the article applies to Excel versions 2019, 2016, 2013, 2010, and Excel for Mac. I have gathered different symbols you cn copy and use. It adds 28 keys to your compatible Apple device, including number keys, function keys, and document navigation controls. Inserting Math Symbols with Keyboard Shortcuts. The YourType Bluetooth Wireless Keypad from Belkin is a convenient solution for data entry, accounting functions, and other numbers-based operations. If you don't already have one, you can create one right now. Try a different character or try to insert it in a different application. - David Carlisle Mar 12 '14 at 1:01. If you like this Page, please click that +1 button, too. Wingdings Checkmark or Tick Symbol. Then you'll see tons of symbols you can attach to your vowels, including the long vowel and short vowel symbols. In this post, I am gonna show you how to write Mathematic symbols in markdown. ⌥ N and ⇧ A. Greek Letter Name. Made up of parts or qualities that are disparate or otherwise markedly lacking in consistency:. Check out all of. Check the box next to "Show keyboard and emoji viewers in menu bar". 360° = 400 grad. To Activate the Keyboard Viewer: 1. Apple has it's own method and doesn't rely on these codes. )Accents and diacriticals Accent A a E e I i O o U u Acute (´) 0180 Á = 0193 á = 0225 É = 0201 é = 0233 Í = 0205 í = 0237 Ó = 0211 ó = 0243 Ú = 0218 ú = 0250. Definition of corresponding in the Idioms Dictionary. A checkmark (also known as checked, tick or checkbox in English) is a mark ( , , , √) used to indicate the concept of "yes", to indicate that "yes, it has been verified" and also, "yes, that is the correct answer". The topic will print as it appears online. gradians / gons. In general, if you're used to Latex, then you can simply enter Latex codes such as \rightarrow in math mode, and LyX will display most of the symbols correctly (you may have to press the SPACE key or move the cursor before LyX displays the symbol). Free icons of Currency symbol in Material Filled style. a with circumflex. In this article, we provide the complete list alt code shortcuts for Mac to insert symbols. The exponent symbol on a keyboard is known as ^. a with ring above. The Mac app is finally stable enough. right from your keyboard. You will see less than or equal to, greater than or equal to, approximately equal to symbols in the option box. not an element of. Wingdings Checkmark or Tick Symbol. Code then, Alt + x. Indeed, the Special Character flyout palette in Adobe InDesign CS3 is a great shortcut to. To insert Greek letter type Ctrl+G ( Command G on Mac OS) and then type Latin letter mentioned in the table below. Tab (Shift-Tab reverses direction) Alternate columns in Column View. Tech support scams are an industry-wide issue where scammers trick you into paying for unnecessary technical support services. Mathematical Symbols Press/hold Alt key while typing the - Starr. I frequently at work have to create line drawings of a photo provided by a client (attach the jpg in Autocad and trace over main features). One of the c. 2 Binary operators. Yahoo; Get our app. Algebra symbols Algebra symbols. We found evidence for a congruency effect on intentional binding. On older versions of OS X, click the Insert button instead. Alt 0153 ™ Alt Code Symbol Alt 231 τ Alt 0184 ¸ Alt 123 { Alt 0203 Ë Alt 0235 ë Filled Arrows Alt 40 ( Alt 232 Φ Web Marketing Alt 125 } Alt 0204 Ì Alt 0236 ì Alt 16 Alt 41 ) Alt 237 φ by Editing Alt 0205 Í Alt 0237 í Alt 17 Plus or Minus Alt 234 Ω www. Below is the complete list of Windows ALT key numeric pad codes for geometric shape symbols, their corresponding HTML entity numeric character references and, when available, their corresponding HTML entity named character references. Apple's brilliance came from their focus on minimalism. This code arises from reorder and expand the set of symbols. Algebra symbols Algebra symbols. While the keyboard on a Mac looks the same as every other keyboard on the face of the planet (except for the command and option buttons, I suppose), it comes packed with a plethora of hidden symbols that users can make use of. Not all code symbols show on printouts even though they show on your screen. Greek Letter Name. It is a variation on the equals sign, which is an expression of mathematical equality. mathl-alt 4. Click Alt+= again to exit from the equitation. To Activate the Keyboard Viewer: 1. Symbols are used to eliminate the need to write long, plain language instructions to describe calculations and other processes. Finding what you want in this massive list of symbols can be quite difficult. By default, the Symbol dialog box shows emoji at first when it's opened. You can help protect yourself from scammers by verifying that the contact is a Microsoft Agent or Microsoft Employee and that the phone number is an official Microsoft global customer service number. The word pairs were presented for 5 s or until participants responded. Open Microsoft Word and go to the Insert tab. AL Go! Generates a HelloWorld project. Review: Apple's 27-inch iMac with Retina 5K display. To use this symbol press down shift and the number 6 on your keyboard. The default Apple keyboard also has two alt keys on the keyboard layout. 1 Unary operators. See also: Hamburger button. e : è é ê ë ē ė ę. Use the keyboard shortcut - ALT + 30 (you need to hold the ALT key and then press 30 from the numeric keypad of your keyboard). Steps to generate and use Math Symbols. The website contains a collection of useful (for me) tips and tricks regading PC keyboards. Net Press/hold Alt key while typing the numbers on the keypad (with Num Lock on). The Triple Horn of Odin is a Viking symbol made of three interlocking horns representing the three horns in the myth regarding Odin and his quest for the magical mead, Odhroerir/Óðrerir, also known as the Mead of Poetry. Go to definition. Use Alt codes to make text symbols and special characters from your keyboard, or laptop. Click the symbol that you want. Also contains a Symbol-to-ASCII converter - Sortable table list of special ASCII characters and character sets including their name, decimal codes, hexadecimal codes, and HTML entity for HTML 4 and HTML 5 compliant sites. Cmd-1, Cmd-2, Cmd-3, Cmd-4. Go to parent folder. To get the letter, character, sign or symbol "|": ( vertical-bar, vbar, vertical line or vertical slash ) on computers with Windows operating system: 1) Press the "Alt" key on your keyboard, and do not let go. You could use \mathbb {Z} to represent the Set of Integers! improve this answer. If you want to know the keyboard equivalent of a symbol, you will find that hovering over a symbol or button will show you the symbol's code. Windows shortcut. This table contains most frequently used text symbols you can input using alt codes, with their names listed. a : à á â ä æ ã å ā. To be on the sure side you should provide a definition of any relation symbol you don't consider to be common knowledge. Open or Start software keyboard in number or symbol mode. Inserting all the mathematical symbols in Microsoft Word is something that you can do right from your keyboard. This code arises from reorder and expand the set of symbols. It is developed by Chinese software firm Kingsoft and consists of WPS Writer, WPS Presentation, and WPS Spreadsheet. The preferred way of symbolizing "approximately equal to" is to use the ≈ symbol, the ~ symbol, or (in some cases) the ≅ symbol (see note 2). You may also want to visit the Mathematics Unicode characters and their HTML entity. To be on the sure side you should provide a definition of any relation symbol you don't consider to be common knowledge. To create an asterisk on a smartphone or tablet, open the keyboard, go into the numbers (123) section, and then the (#+=) or symbols (sym) section. com Alt 28 ∟ Alt 0206 Ï Alt 0238 î. You can insert special characters in your documents and presentations without having to remember all those Alt-codes by using Google Docs and Slides easy-to-use character insertion tool. With Mac OS X 10. Note: In above steps, I have used Microsoft Word 2007. Fast and free shipping free returns cash on delivery available on eligible purchase. Browse: Home / Inserting almost equal / approximate sign (≈) in an Excel cell I want to add the symbol for almost equal, ie ≈, but haven't been able to find it. I have gathered different symbols you cn copy and use. The Symbol will display, select (normal text) in the box of Font and choose Mathematical Operators in the box of Subset. Not all code symbols show on printouts even though they show on your screen. Free icons of Currency symbol in Material Filled style. Use Alt codes to make text symbols and special characters from your keyboard, or laptop. Equivalence & Proportion Operators. Double-click a symbol to "type" it, drag-and-drop it to a text field, or right-click and select Copy Character Info. 2 Binary operators. Degree symbol can be used in case if we're dealing with angles, or when we need to operate with temperature and use Celsius degree. Alt Key Shortcuts – Symbol Categories. a : à á â ä æ ã å ā. How to Type the Trademark Symbol on a Macbook. Letters symbols. Guide for PC and laptop + full list of Alt codes. Apple Footer. math symbols: Asymptotic ≈ Degree symbol ° Delta: Δ : Division sign ÷ Fraction 1/2: ½ : Fraction 1/4: ¼ : Fraction 3/4: ¾ : Greater than > Greater than or equal ≥ Infinity symbol ∞ Left Angle Bracket 〈 Less than < Less than or equal ≤ Micro: µ : Multiplication sign × not symbol ¬ Ohm sign: Ω : Per Mille (1/1000) sign ‰ Pi. In case, you using newer or older version of MS Office, then again you need to follow the same steps. )Accents and diacriticals Accent A a E e I i O o U u Acute (´) 0180 Á = 0193 á = 0225 É = 0201 é = 0233 Í = 0205 í = 0237 Ó = 0211 ó = 0243 Ú = 0218 ú = 0250. Yahoo; Get our app. ALT Codes for geometric shape symbols. You can use Math AutoCorrect symbols as following type one of the following codes followed by a delimiting term. Click the special character you want to insert into the document. What does corresponds expression mean? Sign up with one click: Facebook; Twitter; Google. 360° = 2π rad. ttf (Click Right Mouse " SAVE TARGET AS") Install to Fonts folder of your System (It is easy. 1, Apple supply a system font called Symbol that has the characters at their Unicode code points. your question appears to be about notation rather than tex so may be off topic, but I'd normally denote the set of integers by \mathbb {Z} (with N being natural numbers rather than integers. So here it is: In windows, apparently you hold down Alt+0176. The easiest way to use the congruent symbol is to copy it from another source that uses the symbol and paste it where you need it.
cmmj2imla9m xo5xz8o51uy6v7 vwfb624y82ct u0428kealksiht 62b3r4n9dv8g8f e4px82ug2y1061 y38gqngwtrsrar vimjr5u518 1z6b3fij4uv5y4b w45lw5dabe4oj5 p6y1qg0guoglfkv 9kpj8p699m k7p3m8u2wh3qu0 bo6ar34hw9z khoqktdjghskb mt8q87j6tz0szrj 1y5hxxadrle4q xitl5ojzrlgc o3h801jv0elp wucrgrbg0wk imdznz8vprpmya lkygnhta4y4k49 9qeij1u0elznuge 1563ix2thdrzsq avt5rusjhho65 dfkdtwbiln7
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2020-07-11 02:35:11
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https://astarmathsandphysics.com/ib-maths-notes/calculus/909-differential-equations.html
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## Differential Equations
A differential equation is any equation with one or more differential terms –or similar.
Givento findwe differentiateTo findwe differentiateWe can differentiate any number of times: if we differentiate y n times, we have
Examples of differential equations include:
If a functionis a solution to the differential equation then we can substitutefor into the equation and obtain an identity.
Example: Show thatsatisfies the equation
Hence the given function satisfies the equation.
Example: Show thatsatisfies the equation
Hence the given function satisfies the equation.
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2018-07-18 02:45:42
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http://mathhelpforum.com/latex-help/120991-latex-title-box.html
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# Math Help - latex in the title box
1. ## latex in the title box
is there a way to apply the latex in the title box
be nice if our equations and such could be the same in the title
as in the post.
no big though just more decorative
2. Put the raw code, it should be sufficient
The title is a kind of summary, so you don't have to put many things.
3. I don't think it'd help much, just be over the top so to speak. You can hover over a title to see some text anyway
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2014-04-21 13:20:41
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https://www.csdn.net/tags/MtjaMg4sNTQ3OTUtYmxvZwO0O0OO0O0O.html
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• What is machine learning？ Machine learning is to find functions automatically.For example, speech recognition, image recognition, playing go, dialogue system… 1.regression prediction 2.binary ...
Machine learning is to find functions automatically.For example, speech recognition, image recognition, playing go, dialogue system…
Regression
Prediction.
Binary Classification
Output is yes or no.
Multiclass Classification
Let the machine do multiple choice questions.The machine selects the correct one in the options.
Generation
Produce structured complex things.Like images,sentences created like a human.How to tell the machine what kind of function to look for？
Supervised Learning
Labeled data.Need to give the machine some labeled data for training.
Function Loss
Judge whether the function is good or bad. The basic idea is as follows: First, the image is given to the function for output, and the output result is compared with the label. The proportion of the correct answer is related to the quality of the function.The smaller the LOSS, the closer to the function we need to find. Next, the machine will automatically select the function with the lowest LOSS value.
Reinforcement Learning
The result will allow the machine to evaluate previous decisions.And alpha go both use supervised learning and reinforcement learning.
Unsupervised Learning
Unlabeled data.What can machine learn from unlabeled data.
Function Search Range
Linear,Network Architecture,like RNN and CNN.
Function Search Method
Other Application
Explainable AI
Tell us why this result is output.
Deliberately modify the input data to confuse the machine, resulting in an error in its selection function.
Network Compression
Compress the data input and put it on the phone.
Anomaly Detection
Enter something completely different,Can you let the machine know that ‘i don’t know’?
Transfer Learning
Training data AND testing data.learn by analogy.BUT when testing data totally different with the training data, what tragety will machine choose.
Meta Learning
Let machine can learn by itself! learning algorithmn.Can you make the machine smarter？create learning algorithmn.
Life-long Learning
Machine life-long learning.Continuous learning.
展开全文
• 02What is Machine Learning, Deep Learning and Structured Learning
• What is Machine Learning Two broad classification Supervised Learning Regression Classification Unsupervised Learning What is Machine Learning?One definition of Machine Learning: “A computer program ...
What is Machine Learning
Classification
Unsupervised Learning
What is Machine Learning?
One definition of Machine Learning: “A computer program is said to learn from experience E with respect to some class of tasks T and performance measure P, if its performance at tasks in T, as measured by P, improves with experience E.”
Supervised Learning
Regression
In a regression problem, we are trying to predict results within a continuous output, meaning that we are trying to map input variables to some continuous function.
Classification
In a classification problem, we are instead trying to predict results in a discrete output. In other words, we are trying to map input variables into discrete categories.
Unsupervised Learning
Unsupervised learning allows us to approach problems with little or no idea what our results should look like. We can derive structure from data where we don’t necessarily know the effect of the variables.
With unsupervised learning there is no feedback based on the prediction results.
展开全文
• What is machine learning? machine learning : experience—->skill The skill is something which can improve the performance of program Why use machine learning For example,how to recognize ...
What is machine learning?
machine learning : experience—->skill
The skill is something which can improve the performance of program
Why use machine learning
For example,how to recognize a tree？
One way is that,we may write down many rules of the tree(the rules can be so many),once we need to judge a object whether tree or not we can use these rules to determine.But in fact ,we didn’t recongize by this way.Just think how a three-year-old baby do this.They can recongnize the tree after seen a lot of tree,they don’t need to write down rules.Because they have seen too many tree so when they meet a new thing they can determine whether it is a really tree.
PC can recognize a tree like a three-years-old baby,they just need some data to deal with just like to see many trees like for training.And it would be more effective than writing rules.
What fileds can machine learning apply to?
For example,some interesting field
Navigating on Mars: we can not make rules in advance for the reason that we haven’t reach Mars before,we need machine to collect data and use these data to maker decision when it arrive to Mars
Speech/visual recognition: the feature of voice and image is hard to catch, it also means makeing rules is difficlut so it’s better to use machine
High-frequency trading marketing: it’s hard for human beings to make decision in only just seconds in trading market but marchine learning can do it.The program just need to be trained by the data earlier in the trading market and it would make decision for the later quickly.
Consumer-targeted marketing : make service for people in a large scale in the case the service is targeted,we can’t do it by artificial method,we need use machine learning
What problems is suitable for machine learning
There are three keys which would determine whether a problem fit to machine learning:
(1)Exists some underlying pattern to be bearning
(2)It’s hard to definite rule
(3)There should be a certain amount of data
What’s the aim of machine learning?
We have some input:x and some output:y==>(x,y)
There exists a underlying function f which make f(x)=y,but we don’t know the function f
So we should find hypothesis g : g ≈$\approx$$\approx$ f,there could be many function g ,but we need to chose the best one
What is the difference between machine learning,data mining,artical intelligence and statistics?
Machine learning: use data to compute hypothesis g ≈$\approx$$\approx$targeted f
Data mining: use huge data to find property that is interesting, when the interesting property is hypothesis ,data mining ==machine learning
Artificical Intelligence : compute something that shows intelligent behavior,machine learning can relaize AI
Statistics :use data to make inferenceabout unknown process,traditionaly statistics values useing math provable result ,but machine learning puts highlight on computing data.Statistics can be used to realize ML,provides tools for ML
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• 一、What is Machine Learning ? Two definitions of Machine Learning are offered. 1、Arthur Samuel described it as: “the field of study that gives computers the ability to learn withou...
开始Machine Learning !
一、What is Machine Learning ?
Two definitions of Machine Learning are offered.
1、Arthur Samuel described it as: “the field of study that gives computers the ability to learn without being explicitly programmed.” This is an older, informal definition.
2、Tom Mitchell provides a more modern definition: “A computer program is said to learn from experience E with respect to some class of tasks T and performance measure P, if its performance at tasks in T, as measured by P, improves with experience E.”
Example:playing checkers
E = the experience of playing many games of checkers
T = the task of playing checkers.
P = the probability that the program will win the next game.( performance measure:性能指标）
In general, any machine learning problem can be assigned to one of two broad classifications:
Supervised learning and Unsupervised learning.
附：入门机器学习的知乎帖子：
https://zhuanlan.zhihu.com/p/29704017
展开全文
• we will try to define what it is and also try to give you a sense of when you want to use machine learning. Even among machine learning practitioners, there isn't a well accepted definition of what i....
• I’v been preparing to learn Machine Learning for a few days, but always being delayed by some stuffs like building my blog. After beautifying the blog for a long time, I pick up my course today. ...
• 新定义：A computer program is said to learn from experience E with respect to *(关于，至于)*some class of tasks T and performance measure P, if its performance at tasks in T, as measured by P, impro.....
• t learning anything and I wanted to know, if you had some further debugging mechanics to see better, what is happening inside.</p><p>该提问来源于开源项目：facebookresearch/votenet</p></div>
• 【转载】https://hips.seas.harvard.edu/blog/2013/02/25/what-is-representation-learning/ Posted in Machine Learning. By Roger Grosse – February 25, 2013 In my last post, ...
• one is to learn by yourself, and the other one is learning by lecturers if we think learning as an approach to eat things from the cook that is an interesting approach to see basically, if you learn ...
• Two definitions of Machine Learning are offered. Arthur Samuel described it as: “the field of study that gives computers the ability to learn without being explicitly programmed.” This is an older, ...
• learning : acquiring skill with experience accumulated from observations. machine learning : acquiring skill with experience accumulated/computed from data ML: an alternative ...
• What is Machine Learning? What is Machine Learning? Two definitions of Machine Learning are offered. Arthur Samuel described it as: “the field of study that gives computers the ability to learn ...
• Deep learning 什么是深度学习？ Deep v.s. Shallow 深度学习的大起大落 世人都道深度好，为啥深度好？ Structured Learning 链接：http://speech.ee.ntu.edu.tw/~tlkagk/courses/MLDS_2015_2/Lecture/Brief%...
• 摘要: 本文是吴恩达 (Andrew Ng)老师《机器学习》课程，第一章《绪论：初识机器学习》中第2课时《什么是机器学习？》的视频原文字幕。...What is machine learning? In this article we will try...
• 1.what is Machine Learning? Learning：通过观察获取技能(skill) observation—> learning —>skill Machine Learning：电脑通过数据获得技能(skill) data—>ML—>skill(improve ...
• * supervised learning: * regression * classification *unsupervised learning
• 1.2 [what is machine learning？] 1.人：observation --> learing --> skill 机器：data --> ML --> improved performance measure /skill 2.什么情况下适合使用机器学习： （1）some 'underlying ...
• Meta learning is Learn to learn based on tasks (machine learning is learn to do a given task based on dataset). It does mean that it can handle new tasks with existing status, but it is likely to ...
• 转自：https://www.quora.com/What-is-regularization-in-machine-learning Regularization is a technique used in an attempt to solve the overfitting[1] problem in statistical models.* First of all, I...
• https://www.growthengineering.co.uk/what-is-game-based-learning/ Gamification and game-based learning have been hot topics in the world of learning and development for some time now, but did you...
• Book: An Introduction to Statistical Learning ...http://www-bcf.usc.edu/~gareth/ISL/这是第二章，简要介绍统计学习中的一些基本概念2.1 What Is Statistical Learning? 假定我们观察到一个定量响应变量
• What is machine learning Informal definition:the field of study that gives computers the ability to learn without being explicitly programmed. Modern definition:A computer program is said to learn fr...
• 中文笔记见：...What is Machine Learning? Two definitions of Machine Learning are offered. Arthur Samuel described it as: "the ...
• https://www.datamachinist.com/reinforcement-learning/part-1-what-is-reinforcement-learning/
...
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2021-04-17 20:55:30
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https://www.physicsforums.com/threads/how-fast-do-changes-in-the-gravitational-field-propagate-comments.831289/
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# Insights How Fast Do Changes in the Gravitational Field Propagate? - Comments
1. Sep 7, 2015
### bcrowell
Staff Emeritus
2. Sep 7, 2015
Staff Emeritus
I'm a little surprised that you mention the van Flandern paper. Why pick that particular piece of crackpottery, from amongst the vast selection out there?
It's true that TvF gets the dynamics completely wrong, as Carlip points out. But he gets something much more fundamental wrong: the sun is a static source of gravity. There is no amount of measurement at different positions - which is what planetary orbits essentially do - that will measure the speed of changes in gravitational fields - because the fields do not change.
3. Sep 7, 2015
### Staff: Mentor
This isn't exactly true: the sun rotates and it is emitting radiation and matter, but the gravitational effects of this are (if the back of my envelope is correct) much too small to measure directly.
More important, however, the sun is not the only source of gravity in the solar system; planets have measurable effects on the motions of other planets. So the overall field in the solar system is not static, and propagation effects could in principle be measured. Just not the way Van Flandern was trying to do it.
4. Sep 7, 2015
Staff Emeritus
That is true, although a) as you say, this is not van Flandern's argument, and b) it is small. Really small. Likely smaller than you think. If the solar system consisted of the sun, Jupiter in a circular orbit, and a test mass Earth, there would be no effect. As it is, it's not obvious to me whether the dominant effect is Jupiter's eccentricity or Saturn, although I am leaning towards Saturn.
5. Sep 7, 2015
### bcrowell
Staff Emeritus
I wrote this a long time ago for our FAQ. Greg has copied it to the blog. Most likely I wrote it because somebody was posting about van Flandern at that time.
If the rotation is taken into account, then I think the correct term would be stationary rather than static.
6. Sep 7, 2015
### Greg Bernhardt
Feel free to add on or update it :)
7. Sep 7, 2015
### Staff: Mentor
Considering the sun only, yes. Once the planets are included, in principle I don't think the overall spacetime geometry is even stationary, since the planets aren't in perfectly circular orbits all in the same plane. But it's still very close to stationary in practice (and in turn the effects of rotation are small so "stationary" here is pretty close to "static" in practice).
8. Sep 9, 2015
### john baez
Van Flandern used to post to sci.physics a lot, back when I read that group... forcing Carlip to write a rebuttal at one point. I wonder what became of the guy.
9. Sep 9, 2015
### bcrowell
Staff Emeritus
He died in 2009: https://en.wikipedia.org/wiki/Tom_Van_Flandern
BTW, nice to see you on physicsforums. In discussions here over the years, we've often referred to your online articles and FAQs.
10. Sep 9, 2015
### john baez
Thanks!
Now that van Flandern isn't around to promote his theories on gravity, it might be good to retire him from your FAQ.
It would be interesting to compare that FAQ to this one:
They may both have their own advantages. Maybe someone should combine them. (I know, it's better to do things than suggest that other people do them.)
11. Sep 10, 2015
### bcrowell
Staff Emeritus
As suggested, I've deleted the material about van Flandern.
12. Sep 10, 2015
### john baez
Thanks! Sometimes fighting misconceptions only keeps them alive.
13. Sep 10, 2015
### Jeff Rosenbury
I'm a little unclear here. Why would gravity waves travel at the speed of light?
Light waves travel at the speed of light. There's permittivity and permeability which relate to photons. Assuming the same quantities (or exactly equivalent quantities) apply to gravity seems an odd assumption. It's not even clear to me the dimensions would be the same since the three dimensions seem to fall out of U(1). (Nearly everything we observe comes to us through light. Perhaps that gives us a bias?)
BTW, this isn't intended as criticism. I'm well outside my field (EE) and am only seeking understanding.
14. Sep 10, 2015
### bcrowell
Staff Emeritus
The c in relativity isn't the speed of light. It's a conversion factor between units of time and units of distance. It just also happens to be the speed at which massless things are required to travel according to relativity, and light happens to be a massless thing. So are gravity waves.
All of this is in a vacuum. We're not interested in the index of refraction.
15. Sep 10, 2015
### Staff: Mentor
Gravity waves and electromagnetic waves are both predicted by theories (general relativity for one, Maxwell's electrodynamics for the other) that don't make an allowance for the velocity of the observer, so predict that the speed of the radiation will be the same for all observers. Both theories predict that the propagation speed will be $c$, even though the calculation behind that prediction is different (permittivity and permeability for one, structure of spacetime for the other).
Now, it might seem an amazing coincidence that both calculations yield the same invariant speed - it's not surprising that both speeds are invariant, but why should they be the same? However, it can be mathematically proven that there can be at most one invariant speed. Thus, whatever that speed is, they both have to move at that speed. It's something of a historical accident that we call that invariant speed "the speed of light"; - we discovered light, we measured its speed, and we naturally called the result of those measurements "the speed of light".
Thus, you're thinking about it backwards (although in the order of historical discovery) when you say that the speed of light is what it is because of the values of the vacuum permittivity and permeability. Instead, we should say that whatever units you choose will assign a numerical value to the invariant speed; and that in turn will tell you what the values of the permittivity and permeability must be in that system of units.
Last edited: Sep 10, 2015
16. Sep 25, 2015
### Staff: Mentor
Thank you for the Insights article. The more of those I read, the more I learn.
I'm curious about a corralary question on gravity waves. Several threads disuss gravity wave detection, including the LIGO experiment. Those discussions repeatedly say that detection is more difficult for very distant events. That suggests that gravity waves must dissipate or disperse; is that correct? Where does the energy go?
Your article said, "GR predicts that a gravitational-wave pulse propagating on a background of curved spacetime develops a trailing edge that propagates at less than c.[MTW, p. 957] " That sounds like dissipation; correct?
17. Sep 25, 2015
### john baez
It's also more difficult to see a flashlight when it's miles away than when it's close. The reason is exactly the same: when energy spreads out, there's less of it in any one place. The energy doesn't "go away", since the total amount of energy remains the same. It just spreads out.
It's on my website but it's not my article: if you look at the top of the article you'll see who wrote it.
Nonetheless, I agree with everything in there.
This is called dispersion, which is different than dissipation. Dispersion occurs when waves of different wavelengths move at different speeds. You may have noticed this with sound in air, if you've listened to distant thunder. If you click the link you can read how dispersion of light lets a prism split light into different colors.
Last edited: Sep 25, 2015
18. Sep 26, 2015
### Staff: Mentor
Ay ay! I asked that question stupidly. Apologies. What I was curious about was whether there is a GR term that dampens (dissipates) gravity waves with time, and if yes where the energy goes.
19. Sep 26, 2015
### john baez
No, there's no such term. Gravitational waves don't get dampened and don't lose energy when moving through empty space: they just spread out. In many ways, including this, they behave similarly to electromagnetic radiation in the vacuum.
By the way, the right term is "gravitational waves". "Gravity waves" are something much more familiar:
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2018-05-25 21:18:52
|
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https://theory.eecs.northwestern.edu/DSOM-partial-identification-and-approximation/
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# Online Markets Workshop: Partial Identification and Approximation
The Special Quarter on Data Science and Online Markets will host several workshops on topics related to the special quarter. These workshops are jointly organized with local and visiting faculty in economics, but are targeted to an general audience that is familiar with current research in game theory and algorithms, but may not be familiar with the specific research areas of the workshops.
#### Synopsis
Identification is a fundamental concept in econometrics that ensures that available data is sufficient to recover an economic model. In many cases multiple possible values of model parameters may fit a given distribution of the data equally well. In this case the model is referred to as partially identified. Many important models have been found to be partially identified that includes models of discrete games and auctions. Partial identification is fundamentally tied to the idea of approximation in Computer Science in providing universal bounds for components of an Economic model such as auction welfare or revenue.
Econometric theory has made significant progress in analysis of partially identified models and proposed approaches for their inference. This workshop will introduce basic concepts from econometrics for partially identified models and show the audience recent advances in that literature. The speakers are Ivan Canay (Northwestern), Denis Chetverikov (UCLA), Tatiana Komarova (London School of Economics), and Andres Santos (UCLA)
The technical program of this workshop is organized by Ivan Canay and Denis Nikepelov.
#### Logistics
• Date: Thursday-Friday, April 19-20, 2018.
• Location: Kellogg Global Hub 4101, (map), Northwestern U, 2211 Campus Dr, Evanston, IL 60208.
• Transit: Noyes St. Purple Line (map).
• Parking: Validation for North Campus Parking Garage (map) available at workshop.
• Registration: none necessary, bring your own name badge from past conference.
#### Schedule
Day 1: April 19: Tutorial (Kellogg Global Hub 2410)
• 11:00-12:30: Ivan Canay:
Part I: Introduction to Partial Identification and Inference
• 12:30-2:00: Lunch
• 2:00-3:30: Ivan Canay:
Part II: Introduction to Partial Identification and Inference
Day 2: April 20: Research Talks (Kellogg Global Hub 4101)
• 9:00-9:25: Continental Breakfast
• 9:25-9:30: Opening Remarks
• 9:30-10:10: Andres Santos:
Inference on Directionally Differentiable Functions
• 10:10-10:20: Andres Santos Q/A
• 10:20-10:40: Coffee Break
• 10:40-11:20: Tatiana Komarova:
Binary Choice Models with Discrete Regressors: Identification and Misspecification
• 11:20-11:30: Tatiana Komarova Q/A
• 11:30-12:10: Denis Chetverikov:
Testing Many Moment Inequalities
• 12:10-12:20: Denis Chetverikov Q/A
• 12:20-1:30: Lunch
#### Titles and Abstracts
Speaker: Andres Santos (UCLA)
Title: Inference on Directionally Differentiable Functions
Abstract: This paper studies an asymptotic framework for conducting inference on parameters of the form $\phi(\theta_0)$ where $\phi$ is a known directionally differentiable function and $\theta_0$ is estimated by $\hat{\theta}_n$. In these settings, the asymptotic distribution of the plug-in estimator $\phi(\hat{\theta}_n)$ can be readily derived employing existing extensions to the Delta method. We show, however, that the “standard” bootstrap is only consistent under overly stringent conditions — in particular we establish that differentiability of $\phi$ is a necessary and sufficient condition for bootstrap consistency whenever the limiting distribution of $\hat{\theta}_n$ is Gaussian. An alternative resampling scheme is proposed which remains consistent when the bootstrap fails, and is shown to provide local size control under restrictions on the directional derivative of $\phi$. We illustrate the utility of our results by developing a test of whether a Hilbert space valued parameter belongs to a convex set — a setting that includes moment inequality problems and certain tests of shape restrictions as special cases.
Speaker: Tatiana Komarova (London School of Economics)
Title: Binary Choice Models with Discrete Regressors: Identification and Misspecification
Abstract: In semiparametric binary response models, support conditions on the regressors are required to guarantee point identification of the parameter of interest. For example, one regressor is usually assumed to have continuous support conditional on the other regressors. In some instances, such conditions have precluded the use of these models; in others, practitioners have failed to consider whether the conditions are satisfied in their data. This paper explores the inferential question in these semiparametric models when the continuous support condition is not satisfied and all regressors have discrete support. I suggest a recursive procedure that finds sharp bounds on the components of the parameter of interest and outline several applications, focusing mainly on the models under the conditional median restriction, as in Manski (1985). After deriving closed-form bounds on the components of the parameter, I show how these formulas can help analyze cases where one regressor’s support becomes increasingly dense. Furthermore, I investigate asymptotic properties of estimators of the identification set. I describe a relation between the maximum score estimation and support vector machines and also propose several approaches to address the problem of empty identification sets when a model is misspecified.
Speaker: Denis Chetverikov (UCLA)
Title: Testing Many Moment Inequalities
Abstract: This paper considers the problem of testing many moment inequalities where the number of moment inequalities, denoted by $p$, is possibly much larger than the sample size $n$. There are variety of economic applications where the problem of testing many moment inequalities appears; a notable example is a market structure model of Ciliberto and Tamer (2009) where $p= 2^{m+ 1}$ with $m$ being the number of firms. We consider the test statistic given by the maximum of $p$ Studentized (or $t$-type) statistics, and analyze various ways to compute critical values for the test statistic. Specifically, we consider critical values based upon (i) the union bound combined with a moderate deviation inequality for self-normalized sums,(ii) the multiplier and empirical bootstraps, and (iii) two-step and three-step variants of (i) and (ii) by incorporating selection of uninformative inequalities that are that are far from being binding and novel selection of weakly informative inequalities that are potentially binding but do not provide first order information. We prove validity of these methods, showing that under mild conditions, they lead to tests with error in size decreasing polynomially in $n$ while allowing $p$ while allowing $p$ much larger than $n$; indeed $p$ can be of order $\exp(n^c)$ for some $c>0$. Moreover, when $p$ grows with $n$, we show that all of our tests are (minimax) optimal in the sense that they are uniformly consistent against alternatives whose “distance” from the null is larger than the threshold $(2(\log\,p)/n)^{1/2}$, while {\em any} test can only have trivial power in the worst case when the distance is smaller than the threshold. Finally, we show validity of a test based on block multiplier bootstrap in the case of dependent data under some general mixing conditions.
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2018-10-21 22:57:57
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https://math.tecnico.ulisboa.pt/seminars/tqft/?action=show&id=4483
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# Seminário de Teoria Quântica do Campo Topológica
### Invariants and TQFTs for cut cellular surfaces from finite groups and $2$-groups
We introduce the notion of a cut cellular surface (CCS), being a surface with boundary, which is cut in a specified way to be represented in the plane, and is composed of $0$-, $1$- and $2$-cells. We obtain invariants of CCS's under Pachner-like moves on the cellular structure, by counting colourings of the $1$-cells with elements of a finite group $G$, subject to a “flatness” condition for each $2$-cell. These invariants are also described in a TQFT setting, which is not the same as the usual $2$-dimensional TQFT framework. We study the properties of functions which arise in this context,associated to the disk, the cylinder and the pants surface, and derive general properties of these functions from topology. One such property states that the number of conjugacy classes of $G$ equals the commuting fraction of $G$ times the order of $G$.
We will comment on the extension of these invariants to 2-groups and their (higher) gauge theory interpretation.
This is work done in collaboration with Diogo Bragança (Dept. Physics, IST).
#### Ver também
https://arxiv.org/abs/1512.08263
Organizadores correntes: Roger Picken, Marko Stošić.
Projecto FCT PTDC/MAT-GEO/3319/2014, Quantization and Kähler Geometry.
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2018-04-24 18:06:07
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https://math.stackexchange.com/questions/3121989/simplex-algorithm-determining-two-phase-is-required-and-choice-of-artificial-va
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# Simplex Algorithm, determining Two Phase is required and choice of artificial variables
Given the following system :
\begin{align*} \text{minimise } z = &2x_1 &+ 3x_2 &+ 3x_3 &+ x_4 &- 2x_5& \\ \end{align*}
Subject to \begin{align*} & x_1 &+ 3x_2 & &+x_4 &+ x_5 &= 2 \\ & x_1 &+ 2x_2 & &- 3x_4 &+ x_5 &= 2 \\ - &x_1 &- 4x_2 & +x_3 & & &= 1 \\ \end{align*}
with $$x_1, x_2, x_3, x_4, x_5 \geq 0$$
There should be Phase I and then Phase II of the simplex method.
### Q2 - how to know which rows should have artificial variables added
For question 1, the objective function can be written as
\begin{align*} -z + 2x_1 + 3x_2 + 3x_3 + x_4 - 2x_5 = 0\\ \end{align*}
The way that the system is initially set up has basic variables $$x_3$$ and non-basic variables $$x_1,x_2,x_4,x_5 = 0$$.
Meaning the objective function is
\begin{align*} -z + 3x_3 = 0\\ \end{align*}
Or
\begin{align*} z = 3x_3 \end{align*}
Why is this an issue?
For Question 2 I'm not sure what to consider.
• 1. You need 3 basic variables to start; 2. Your starting point does not satisfy the constraints. 3. Is there no nonnegativity constraint? – LinAlg Feb 22 at 1:50
• @LinAlg (1) why do I need three variables to start (because that's the dimension of the row-space and solution?) (2) what do you mean? That my starting point is $x_3 = 1$, and if all other $x_i=0$ the second (and first) equalities (constraints) are false (3) There is - I'll add that – baxx Feb 22 at 2:30
• 1. since you have 3 constraints; 2. 'are false': that is not ok. You need basic feasible solutions in every iterate (so basic AND feasible; you have neither) – LinAlg Feb 22 at 2:46
Point (2) in OP's comment is very near to the an answer.
For each of the first two constraints, there's no decision variable
• which has nonzero coefficient, and
• which doesn't appear in the other two rows.
For a concrete counterexample, you may consider the third row, in which $$x_3$$
• has coefficient one
• doesn't appear in the first two constraints.
Therefore, in general, the first two rows need artificial coefficients.
Remarks: Observe that $$\begin{vmatrix} 1 & 3 & 0 \\ 1 & 2 & 0 \\ -1 & -4 & 1 \end{vmatrix} \ne 0$$ and $$\begin{vmatrix} 0 & 1 & 1 \\ 0 & -3 & 1 \\ 1 & 0 & 0 \end{vmatrix} \ne 0$$, so you may actually solve it as in in exams/tests so save the trouble of doing phase I. However, finding initial BFS like this is never a general method.
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2019-06-19 21:49:09
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https://www.stsci.edu/hst/instrumentation/acs/performance/photometric-cte-corrections
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## Introduction
Since 2003, the ACS Team has taken observations of 47 Tuc to quantify the dependece of stellar photometry on the number of parallel and serial transfers. The observations employ a specialized dither pattern that allows one to derive an analytical model that depends on the following parameters: stellar flux, background level, and the number of parallel transfers. Losses due to serial transfers are currently very small (< 2% at the edge of the chip and far from the amplifiers, where losses are the worst) and consistent with zero for the level of sky background usually achieved by GO observations (> a few electrons per pixel). In 2012, the original model was revisited and the assumption of a log-space linear relationship between the magnitude losses as a function of stellar flux (valid for the small CTE losses in pre-SM4 data) was shown to be an oversimplification of the CTE effects after ~10 years in space. This led to a new and improved model that also correctly accounts for the time-dependence of the derived coefficients (see ACS ISR 2012-05).
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2020-07-12 16:57:21
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http://math.stackexchange.com/questions/335938/property-of-sequence-of-eigenvalues-of-an-operator
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# Property of sequence of eigenvalues of an operator.
For a positive (self adjoint) operator $A$ with eigenvalues $\lambda_k$, is it possible to have the case when neither $\lambda_k\to \infty$ or $sup_k \lambda_k<\infty$ for example if a subsequence tends to $\infty$ and another subsequence stays bounded? In a paper I am reading (Continuity of $l^2$-valued Ornstein-Uhlenbeck Processes), they only check 2 cases.
If so can one order the eigenvectors so that it is not the case?
Thanks.
-
So you mean $\limsup \lambda_k=+\infty$ and $\liminf \lambda_k<\infty$. – 1015 Mar 20 '13 at 15:45
Yes. That's right. – David Mar 20 '13 at 15:47
The operator will not be bounded, as soon as $\limsup \lambda_k=+\infty$ (equivalently $\sup \lambda_k=+\infty$). Take a subsequence $\lambda_{n_k}$ tending to $+\infty$. For each $k$ take a norm $1$ eigenvector $x_k$ associated with $\lambda_{n_k}$. Then $$\|A\|\geq \|Ax_k\|=\lambda_{n_k}\longrightarrow +\infty.$$ The fact that $\liminf \lambda_k<\infty$ will not change that. Neither the fact that $A$ is positive, self-adjoint, or whatever.
But you can construct unbounded examples, of course. It suffices to take the diagonal operator $\mbox{diag}(\lambda_k)$ in any orthonormal basis. Then the domain is the susbspace of all vectors $x=(x_k)$ such that $\sum_k \lambda_k^2x_k^2$ converges. This will depend on the sequence, but of course it will always contain the vectors with finitely many nonzero coordinates.
I am not sure I understand what you are saying. The operator $A$ is allowed to be unbounded. – David Mar 20 '13 at 15:56
@David What do you mean, order the eigenvectors? That's like reordering the $\lambda_k$'s. But you still have $\limsup =+\infty$ and $\liminf<\infty$. So permute the elements of the diagonal accordingly. This is still an example. – 1015 Mar 20 '13 at 16:48
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2016-07-23 13:39:38
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https://perturbo-code.github.io/mydoc_tutorial_gaas.html
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Example calculation in a polar material with long-range e-ph interactions.
Run the calculations in the directory “pw-ph-wann” to obtain the data needed to run qe2pert.x. Run qe2pert.x to get ‘prefix’_epwan.h5, which is required for all calculations using perturbo.x.
The input file for each calculation using perturbo.x is similiar to the silicon case (“example02-silicon-perturbo”, link). The main difference is the 'imsigma' calculation, where users can use a variable called polar_split to specify whether they want to compute the full matrix element (polar plus non-polar part), or just the polar or nonpolar part.
• For the long-range (polar) part, we set polar_split='polar' in the input file. In this example, we use 'cauchy' for $$\mathbf{q}$$ point importance sampling and set the variable cauchy_scale for the Cauchy distribution.
• For the short-range (nonpolar) part, we use rmpolar for the variable polar_split and 'uniform' for sampling.
Remember to converge both the long- and short-range parts of the e-ph matrix elements with respect to the number of $$\mathbf{q}$$ points (the variable nsamples). If polar_split is not specified, perturbo.x will compute e-ph matrix elements including both the short- and long-range interactions, which typically has a slow convergence with respect to number of $$\mathbf{q}$$ points.
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2022-08-07 18:42:32
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https://en.wikipedia.org/wiki/Pentation
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# Pentation
In mathematics, pentation is the next hyperoperation after tetration but before hexation. It is defined as iterated (repeated) tetration, just as tetration is iterated exponentiation.[1] It is a binary operation defined with two numbers a and b, where a is tetrated to itself b times.
## History
The word "pentation" was coined by Reuben Goodstein in 1947 from the roots penta- (five) and iteration. It is part of his general naming scheme for hyperoperations.[2]
## Notation
There is little consensus on the notation for pentation; as such, there are many different ways to write the operation. However, some are more used than others, and some have clear advantages or disadvantages compared to others.
• Pentation can be written as a hyperoperation as ${\displaystyle a[5]b}$. In this format, ${\displaystyle a[3]b}$ may be interpreted as the result of repeatedly applying the function ${\displaystyle x\mapsto a[2]x}$, for ${\displaystyle b}$ repetitions, starting from the number 1. Analogously, ${\displaystyle a[4]b}$, tetration, represents the value obtained by repeatedly applying the function ${\displaystyle x\mapsto a[3]x}$, for ${\displaystyle b}$ repetitions, starting from the number 1, and the pentation ${\displaystyle a[5]b}$ represents the value obtained by repeatedly applying the function ${\displaystyle x\mapsto a[4]x}$, for ${\displaystyle b}$ repetitions, starting from the number 1.[3][4] This will be the notation used in the rest of the article.
• In Knuth's up-arrow notation, ${\displaystyle a[5]b}$ is represented as ${\displaystyle a\uparrow \uparrow \uparrow b}$ or ${\displaystyle a\uparrow ^{3}b}$. In this notation, ${\displaystyle a\uparrow b}$ represents the exponentiation function ${\displaystyle a^{b}}$ and ${\displaystyle a\uparrow \uparrow b}$ represents tetration.The operation can be easily adapted for hexation by adding another arrow.
• Another proposed notation is ${\displaystyle {_{b}a}}$, though this is not extensible to higher hyperoperations.[6]
## Examples
The values of the pentation function may also be obtained from the values in the fourth row of the table of values of a variant of the Ackermann function: if ${\displaystyle A(n,m)}$ is defined by the Ackermann recurrence ${\displaystyle A(m-1,A(m,n-1))}$ with the initial conditions ${\displaystyle A(1,n)=an}$ and ${\displaystyle A(m,1)=a}$, then ${\displaystyle a[5]b=A(4,b)}$.[7]
As its base operation (tetration) has not been extended to non-integer heights, pentation ${\displaystyle a[5]b}$ is currently only defined for integer values of a and b where a > 0 and b ≥ −1, and a few other integer values which may be uniquely defined. Like all other hyperoperations of order 3 (exponentiation) and higher, pentation has the following trivial cases (identities) which holds for all values of a and b within its domain:
• ${\displaystyle 1[5]b=1}$
• ${\displaystyle a[5]1=a}$
• ${\displaystyle a[5]0=1}$
• ${\displaystyle a[5](-1)=0}$
Other than the trivial cases shown above, pentation generates extremely large numbers very quickly such that there are only a few non-trivial cases that produce numbers that can be written in conventional notation, as illustrated below:
• ${\displaystyle 2[5]2=2[4]2=2^{2}=4}$
• ${\displaystyle 2[5]3=2[4](2[4]2)=2[4]4=2^{2^{2^{2}}}=2^{2^{4}}=2^{16}=65,536}$
• ${\displaystyle 2[5]4=2[4](2[4](2[4]2))=2[4](2[4]4)=2[4]65536=2^{2^{2^{\cdot ^{\cdot ^{\cdot ^{2}}}}}}{\mbox{ (a power tower of height 65,536) }}\approx \exp _{10}^{65,533}(4.29508)}$ (shown here in iterated exponential notation as it is far too large to be written in conventional notation. Note ${\displaystyle \exp _{10}(n)=10^{n}}$)
• ${\displaystyle 3[5]2=3[4]3=3^{3^{3}}=3^{27}=7,625,597,484,987}$
• ${\displaystyle 3[5]3=3[4](3[4]3)=3[4]7,625,597,484,987=3^{3^{3^{\cdot ^{\cdot ^{\cdot ^{3}}}}}}{\mbox{ (a power tower of height 7,625,597,484,987) }}\approx \exp _{10}^{7,625,597,484,986}(1.09902)}$
• ${\displaystyle 4[5]2=4[4]4=4^{4^{4^{4}}}=4^{4^{256}}\approx \exp _{10}^{3}(2.19)}$ (a number with over 10153 digits)
• ${\displaystyle 5[5]2=5[4]5=5^{5^{5^{5^{5}}}}=5^{5^{5^{3125}}}\approx \exp _{10}^{4}(3.33928)}$ (a number with more than 10102184 digits)
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2018-12-19 01:46:16
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https://codereview.stackexchange.com/questions/94214/number-of-sub-strings-with-same-first-and-last-character
|
# Number of sub strings with same first and last character
Problem Statement:
You are given a function $f(x)$, where $f(x)$ is $1$ if the first and last characters of string $x$ are equal; else it is $0$. You are given a string $S$ and you have to find the sum of $f(x)$ for all substrings $x$ of given string $S$.
Sample Input:
7
ababaca
Sample Output:
14
Explanation:
f("a")=1, f("aba")=1, f("abaca")=1 but f("ab")=0, f("bac")=0. Hence counting all substrings we get 14.
The 14 substring are
a - 4(times)
b - 2
c - 1
aba - 2
bab - 1
aca - 1
ababa - 1
abaca - 1
ababaca - 1
My code:
l = int(input())
s = [x for x in input().strip()]
print(l + sum([1 for i in range(0,l-1) for j in range(i+1,l) if s[i] == s[j]]))
I have tried the following snippets too:
Timeout:
from collections import Counter
import math
l, s = int(input()), [x for x in input()]
print(l + sum([math.factorial(value-1) for value in Counter(s).values() if value != 1]))
import itertools
l, s = int(input()), input()
print(sum(int(i[0] == i[-1]) for i in itertools.combinations_with_replacement(s,2)))
The solutions is working fine except that it times out in certain test cases. Can anyone suggest a better way to do this?
First review
It is almost always a good idea to put your logic in a function (or class) that can be easily documented and tested. That way, in anything goes wrong during the optimisation, you'll know it straight-away. Also, this can be useful if you need to measure your result to ensure your optimisation is an actual optimisation.
Just moving the code around, here is what I start with :
def get_nb_substring(s):
l = len(s)
return(l + sum([1 for i in range(0,l-1) for j in range(i+1,l) if s[i] == s[j]]))
def test_from_input():
_ = int(input()) # useful ?
s = [x for x in input().strip()]
print(get_nb_substring(s))
def automatic_tests():
assert get_nb_substring("ababaca") == 14
if __name__ == '__main__':
automatic_tests()
Now, the actual review may start.
• you don't need 0 as a first argument for range.
• you usually don't need the length when you are using iterations in Python. When you do need it, it usually means you are doing something wrong.
• for that reason, what you are doing with indices can be achieved with itertools.combinations(s, 2)
Now, the code can be rewritten :
def get_nb_substring(s):
return len(s) + sum(1 for i, j in itertools.combinations(s, 2) if i == j)
which is still 0(n^2).
Optimisations
An idea could be to check where the different letters appear. Then if a letter c appears at positions p1, p2, ..., pn, you know it will contribute for substrings : p1, p1-p2, p1-p3, ..., p1-pn, p2,p2-p3, p2-p4, ... p2-pn, .... pn. There will be 1+2+...+n = n*(n+1)/2 such substrings.
There, if I did everything correctly, I end up with O(n) code :
def get_nb_substring(s):
position = dict()
for i, c in enumerate(s):
position.setdefault(c, []).append(i)
return sum(l*(l+1)/2 for l in (len (l) for l in position.values()))
On the test example, it seems to work. On other cases, I obtain different results than your code so I don't know which is correct.
I've used a dict to map characters to a list of position but a mapping from characters to number of positions would have been enough. This corresponds to using the Counter collection)
def get_nb_substring(s):
return sum(v*(v+1)/2 for v in collections.Counter(s).values())
If you append some character c to a string, how many new substrings can you make that start with c and end with c? As many as the number of c you have seen so far.
You can iterate over the characters of the input string and build a map of counts. Before incrementing the count for a character, add to the running sum the current count. In case of ababaca, this will happen:
• a : current count 0, set it to 1
• b : current count 0, set it to 1
• a : current count 1, add it to the sum, increment count
• b : current count 1, add it to the sum, increment count
• a : current count 2, add it to the sum, increment count
• c : current count 0, set it to 1
• a : current count 3, add it to the sum, increment count
The total sum at this point is 7, add to this the number of characters. Time complexity $O(N)$
Btw, there's no need to make a list of characters like this:
s = [x for x in input().strip()]
If you do just this your program will work just the same:
s = input().strip()
Putting it together:
from collections import Counter
def count_begin_end(word):
total = len(word)
counter = Counter()
for c in word:
total += counter[c]
counter[c] += 1
def main():
_, word = input(), input()
print(count_begin_end(word))
if __name__ == '__main__':
main()
You're creating a few lists where you don't need them. I don't know if this fixes the timeout problem, but here's my take:
def countF(s):
l=len(s)
return sum(int(s[i]==s[j-1]) for i in range(l) for j in range(i+1,l+1))
print(countF("ababaca")) #14
First of all, I'm not transforming the string into a list — it's not needed, you can get to a specific characters using square brackets just like you could in a list.
Second, instead of a list comprehension, I'm using a generator comprehension — this way no list is built up in memory. For example, given a very long string, your code needs to build a list the size of the cardinality of possible substrings, which can get quite large. Using my generator, it sums up while looping, so it's always just a single number it needs to keep in memory.
edit: Good point about the slicing, this way should be better.
• Hey.. Appreciate your reply. But can you explain how this solution is different to mine? – Gaurav Keswani Jun 20 '15 at 22:11
• Also, wouldn't s[i:j] create a substring which 1) Is an expensive command 2) is unneeded since we only want to compare the first and last character of the substring? – Gaurav Keswani Jun 20 '15 at 22:12
• @GauravKeswani Was just editing it. Is this enough? – L3viathan Jun 20 '15 at 22:12
• The generator comprehension bit makes sense to me. But can you explain why we are creating the substring? It isn't needed right? – Gaurav Keswani Jun 20 '15 at 22:13
• Good point, my edit should make it faster. int isn't really needed, but I doubt it makes it any slower either. – L3viathan Jun 20 '15 at 22:14
Since this is something totally different, I have another answer, which should be in $O(N)$. With standard lib, but all unoptimized at the moment and just proof of concept:
from collections import Counter
s="ababaca"
i=7
c=Counter()
for ch in s:
c[ch]+=1
n+=c[ch]-1
print(n+i)
• I was thinking on similar lines and had come up with this some time back from collections import Counter import math l, s = int(input()), [x for x in input()] print(l + sum([math.factorial(value-1) for value in Counter(s).values() if value != 1])) – Gaurav Keswani Jun 20 '15 at 22:36
After numerous suggestions on this portal and a good night's sleep, I could come up with this :)
from collections import Counter
_, s = int(input()), input()
occurences = Counter(s).values()
print(sum(n*(n+1)//2 for n in occurences))
And it works! :D
• Is this an O(1) solution or a O(NlogN) solution? I think it is nlogn but I need confirmation. – Gaurav Keswani Jun 21 '15 at 20:13
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2019-12-13 04:03:37
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https://frost3d.ru/eng/faq-working-in-software/
|
Change language:
### 1. Installation/reinstallation/licensing/start errors
1. Q: There are errors of connection with the server at the software start or at computation start.
A: As a rule, errors of connection with the server are connected with problems/changes in network infrastructure of the user and/or access rights of the user account.
1. It is necessary to provide access to the following network nodes:
IP :port 109.120.178.199 :1375 109.120.178.199 :8733 188.227.72.106 :1375 188.227.72.106 :8733
1. Specify/renew settings (login and password) of proxy server if it’s used for Internet connection.
Specify settings of proxy server in file NetUtil.exe, automatic address: %SYSTEMDRIVE%\Program Files\Simmakers\Frost 3D\NetUtil.exe.
For reliable connection test with Frost 3D servers it is necessary to execute telnet inquiries to the listed addresses and ports by means of the standard telnet-client of Windows.
If the attempt of connection leads to error, then, most likely, access to Frost 3D servers is adjusted incorrectly (necessary ports and addresses are blocked by brandmauer or other means of network protection).
[свернуть]
2. Q: At start the message "License verification. License renewal error. Repeat synchronization" or "License verification. License is blocked on the server" appears.
A: This error can be caused by:
1. The problem connected with Frost 3D servers connection (see Q: There are errors of connection with the server at the software start or at computation start).
2. Wrong time zone, inexact time
Check correctness of the time zone settings on the computer. Staticize data and synchronize time.
1. Start of the software without the rights of the Administrator
Close the software. Choose one of the offered options:
• Give the rights of the Administrator to the current user and start Frost 3D on behalf of the current user;
• Start Frost 3D on behalf of the Administrator by means of a context menu;
• Establish a check mark «To start on behalf of the Administrator» in Frost 3D properties (a label, or the executable file) and start the software.
1. The software installation and operating use from different accounts
Change of the user can break the operation mode since some settings and components of the software are saved in the Register and Documents of the user. According to the instruction for installation it is understood that the software installation and operation are carried out by the same user.
1. The software installation without Administrator’s rights
• Delete the software;
• Reinstall Frost 3D with the Administrator’s rights
1. Absence of necessary software components
Check the existence of:
• Microsoft Visual C++ Redistrubutable Package 2013*
• Drivers of USB keys Guardant version 31
*You can download the installers on the official site of MS Upload Center.
Make corresponding changes. In case of the repeated error, reinstall the software.
[свернуть]
3. Q: How to reinstall Frost 3D?
A: For reinstallation of the software:
1. Remove earlier installed version of Frost 3D:
• Start-up menu → uninstallation Label, or
• Control panel«Programs and components» list.
2. Start installers* Net Framework 5.1 and Visual C++ Redistributable Package for checking its earlier installation.
*You can download the installers on the official site of MS Upload Center.
1. Start the installer of the driver of USB keys of Guardant version 6.31.
2. Install the software according to the instruction (see the Instruction for installation).
[свернуть]
4. Q: There were problems with reinstallation of the software on other computer. How is it possible to configure the software anew?
A: At change of the computer configuration address to Technical support $support@simmakers.com$ or by +7 495 772 54 07. After verification of changes, you can start the software installation according to the instruction from the delivery set.
[свернуть]
5. Q: There is an error of reading the default.sta file when opening the "Example of Creation of Model of the Well" project. Is it critical?
A: The error of reading default.sta file is not critical. The absent file does not influence results of computation. Usually this mistake arises when opening the projects created in earlier versions of the software.
[свернуть]
6. Q: What to do if the software began to work more slowly?
A: For productivity increase of the software, we recommend to check:
1. Project settings:
• Filtration computation existence;
• The number of iterations (saved results of computation);
• Size of computational area.
1. Serviceability of the video card and drivers if computation is conducted on the video card. In case of failure, reinstall the driver of NVidia video card.
2. Way of computation saving. Give preference to the internal hard drive, but not portable devices of storage of information (flash memory cards, external hard drives, etc.).
3. General loading of the operating system. Productivity of the software significantly decreases if several resource-intensive processes are started at the same time.
4. Replacement of USB-port to which the USB-key is connected. USB-ports can have different productivity.
5. In rare cases productivity of USB-key can decrease due to malfunction of the driver of USB-keys. Reinstallation of the driver and/or reset of the computer allow to eliminate malfunction.
[свернуть]
7. Q: In certain cases it is not possible to rename computation via the interface of the software (see Figure 1). In attempt to apply changes nothing occurs.
Fig 1 – «Choice of displayed computation» window.
A: The name of computation is the name of the directory in which it is stored. Respectively, restrictions for renaming of computation coincide with restrictions for renaming of directories, for example*:
1. Names of directories are limited on length;
2. Names of directories shouldn’t contain the forbidden signs;
3. Names of directories shouldn’t repeat;
4. Names of directories shouldn’t coincide with the names reserved in Windows (CON, PRN, AUX, CLOCK, NUL, COM1, COM2, …, COM9, LPT1, LPT2, …, LPT9, etc.).
*You can study the complete list of restrictions on the official site of Microsoft: //msdn.microsoft.com/en-us/library/aa365247.
It is necessary to remember that gaps at the end of the name are automatically removed by the operating system.
Important: The software displays the list of the computations which were obtained at the last scanning directories. If necessary, update this list manually, pressing «Reboot Computation» button.
[свернуть]
### 2. Simulation: thermal physics, heat equation, properties and etc
8. Q: : Is it possible to consider temperature changes within one day/hour at a task of dependence of temperature from time in Boundary conditions /External environment? For example: within one day temperature is 0 °C (12:00) and 100 °C (13:00)?
A: The software package allows to consider change of dependences on time with the resolution up to 1 second. Nevertheless, we recommend not to set change of values with a step, smaller than 1 hour.
[свернуть]
9. Q: Is it possible to use date of measurements of ground temperature distribution on depth as the start date of computation (the initial moment of the simulated period)?
A: Yes, as a rule, date of measurements of ground temperature is accepted as the start date of computation.
[свернуть]
10. Q: Is it necessary to set dependences of physical quantities on time in Boundary conditions since start date of the simulation period?
A: Start date of computation does not influence the dependences on time of air temperature, solar radiation or other physical quantities. All physical quantities are «tied» to the calendar and do not depend on start date of computation.
Important: Make sure that all dependences of physical quantities on time are set on the whole simulation period.
[свернуть]
11. Q: Is it possible to use the constant ground temperature (constant temperature along the whole depth, along the whole computational domain) as initial distribution of ground temperatures?
A: For obtaining correct results of simulation, we do not recommend to use constant temperature as initial distribution of temperatures in ground. Considering the fact that ground temperature changes along the depth, it is necessary to use data of thermometric research.
Important: For short period of simulation (1-3 years), we recommend to specify the most exact values of initial ground temperature.
Constant temperature of ground along the depth is used when there is no data of thermometric research or initial distribution of ground temperature is ignored*.
*It is possible to ignore initial distribution of temperatures only in a limited class of long simulation period under natural conditions.
[свернуть]
12. Q: What is «Computation under natural conditions» and why to apply it?
A: «Computation under natural conditions» is a numerical experiment which allows to estimate distribution of ground temperatures, proceeding from climatic data and thermophysical properties of ground. Nevertheless, some institutes use this method for receiving initial distribution of temperature in ground, which is, in our opinion, not always reasonable due to the low accuracy of basic data (climatic data and/or data of thermometry).
We recommend to carry out «Computation under natural conditions» only for check of compliance of climatic data to the thermometry data:
1. Create three-dimensional model of geology without construction objects and technogenic influences;
2. Set climatic data;
3. Set approximate temperature along the whole depth. As a rule, the temperature which is typical for this region is used;
4. Carry out computation for 50-100 years period.
In 50-100 years temperature in ground will be stable, and will correspond to the climatic conditions. The received values of temperatures have to be close to the data of thermometry received experimentally.
[свернуть]
13. Q: After start of the computation it stops and the software says «Memory error».
A: This error arises in case when the software lacks RAM and video RAM for the problem solution. The most common causes of this error are:
1. the user specified value of thermal capacity of one of materials by mistake in «Database of materials and boundary conditions». Users can specify value in MJ/ (m3*k) or in kJ / (m3*k) instead of J / (m3*k) by mistake. Recheck values of volume thermal capacity of each material attentively in «Database of materials and boundary conditions»;
2. the computational mesh contains too large number of nodes (cells). It is possible to reduce number of nodes without loss of accuracy using the markers of the computational mesh. If the markers of computational mesh are already placed in the optimum way (from the user’s point of view), then it is necessary to make the computational mesh more rough;
3. the minimum step on space of computational mesh is extremely small (less than 1 cm);
4. one of materials of the computational mesh has extremely high heat conductivity (more than 1000);
5. very long simulation time period (more than 1000 years);
6. any combination of points 1,3,4,5, in addition complicates a task and narrows the admissible borders specified in the corresponding points. For example, computation for 500 years through a steel wall 1 cm thick, which heat conductivity is equal 48, can cause «Memory error» since problem definition approaches the borders specified in points 3, 4 and 5.
[свернуть]
### 3. Software interface: use of buttons, windows
14. Q: How to consider temperature wells correctly?
Q: I carry out the following sequence of actions:
1. Create the temperature well and specify some dependence of temperature on depth in «Database of materials, physical properties and heat exchange conditions» (hereinafter DB) in the «Thermal Distribution» tab.
«Thermal Distribution» tab Specifying (setting) the thermometric wells.
2. Set ground temperature which differs from the temperatures set for the temperature well in DB in «Materials» tab.
3. Perform computation.
In the computation results there’s no influence of the temperature well.
A: In «Database of materials, physical properties and conditions of heat exchange» (hereinafter DB) only one of two ways of setting initial temperature can be applied to each Material:
a) To use three-dimensional thermal distribution;
b) To use separate tabular dependence of temperatures on depth.
It is connected with the fact that ground cannot have two various values of initial temperature.
Figure 1 – «Materials» tab: indication of initial ground temperature.
For use of temperature wells for the purpose of setting initial ground temperature:
1. Add information on temperature wells to DB in «Thermal Distribution» tab;
2. In DB in «Materials» tab choose «To use thermal distribution» for all materials which temperatures have to correspond to three-dimensional temperature distribution (see Figure 2).
Figure 2 – Switch of thermal distribution.
3.Perform computation.
Temperatures of ground in initial timepoint will correspond to the set (by means of temperature wells) three-dimensional thermal distribution.
[свернуть]
15. Q: Does the difference of temperatures set in boundary condition on cooling device take into account the sign or on the module?
A:The difference of temperatures should be set by positive value.
[свернуть]
16. Q: Why at change of position of one of three-dimensional objects in «3D Editor» and creation of «Computational mesh», the quality of sampling of the moved object became significantly lower?
A: For quality sampling of an object the markers of computtional mesh (markers of splitting and/or condensation) were used. At change of provision of three-dimensional objects, it is also necessary to move markers of computational mesh which will be involved at creation of computational mesh.
[свернуть]
17. Q: How to change the position of markers of computational mesh during «3D Editor» step?
A: The position of markers of computational mesh can be changed only in «2D Editor». For this purpose, pass in «2D Editor» (at any time, before creation of computational mesh) and edit the position of markers. Recreation of three-dimensional model is not required.
[свернуть]
18. Q: How to change the position of markers of computational mesh in «2D Editor»?
A: On the left side panel there is a group of buttons which allow to add new markers (in more detail in «User’s guide», point 3.1.1). Markers can be allocated and edited in the field of the list (the right part of the screen) and on the stage (the field of work with geometry, in the center of the screen). For viewing and changing the properties of markers allocate one or several markers in the field of properties (the right lower part of the screen).
Figure 1 — Properties of mesh marker
Absolute coordinate is a coordinate on Z axis. Pay attention that units of measure of Z coordinate can differ from units of measure of X,Y coordinates. Often the unit of measure for absolute marks differs from the unit of length measure.Thus, in Figure 1 Z coordinate of the absolute mark is measured in meters and X and Y coordinates are measured in millimeters.
Marker type defines a method of impact of a marker on the computational mesh (a marker of splitting or a marker of condensation). Markers of condensation allow to create condensation of computational mesh near their situation. Markers of splitting allow to control manually.
Type of marker positioning defines coordinate axes which the marker will influence.Possible options:
• XYZ
• X
• Y
• Z
Belong to the object – all markers have to belong to «Markers» object.
Coordinates (position) of a marker define a point in which condensation (for condensation markers) or splitting (for splitting markers) the computational mesh will be carried out. For example, if the type of positioning of a marker is set in value X, then condensation or splitting will be carried out only on the set axis. At creation of computational mesh all other coordinates of a marker will be ignored. We recommend to have a look at a bright example of markers with various type of positioning in Figure 2.
Figure 2 – Examples of a condensation marker impact on computational mesh
In Figure 2 markers of condensation of different type of positioning are represented (red color). Blue and green lines designate the mesh (borders of cells of the computational mesh). Black lines designate coordinate axes.
Figure 2a – condensation markers with type of X type positioning. Reduction of a step of the computational mesh on space on the coordinate of X near a marker is observed. Y and Z coordinates of a marker do not influence the computational mesh (the same effect will be reached if to arrange a marker in a position of a red circle).
Figure 2b — condensation markers with Y type of positioning. Reduction of a mesh step on space on Y coordinate near a marker is observed. X and Z coordinates of a marker do not influence the computational mesh (the same effect will be reached if to arrange a marker in a position of a red circle).
Figure 2c is a condensation marker with XYZ positioning type. Reduction of step of the mesh on all coordinates is observed.
Important: to display parameters and/or markers quantity changes on the computational mesh:
• Reconstruct the computational mesh, having pressed the «Following Step» button in «3D Editor»;
• Check parameters of the computational mesh creation.
For detailed information concerning the modes of creation of the computational mesh, we recommend to study the point 4.6 «Start of 3D model transfer on the computational mesh» of the software documentation.
Examples of practical application of markers are stated in free videocourses of Frost 3D training, access to which you can request from the managers of our company.
[свернуть]
19. Q: How to create the table of dependence of temperature on depth for several timepoints?
Q: The example of the necessary scheme is represented in Figure 1.
Figure 1 — Example of table for report.
A: Frost 3D gives an opportunity of automatic creation of the table of dependences of temperature on depth at the same time for several timepoints. For creation of the scheme of the specified format (see Figure 1), we recommend to use Schedule of temperature along a vertical (the thermometric well).
Pay attention to the following designations:
$t_{1},t_{2},…,t_{n}$ timepoints for which it is necessary to receive dependence of temperatures on depth; $n$ number of timepoints $t_{i}$; $V_{x,y}$ the vertical line in three-dimensional space, coordinates in the horizontal plane of which – $(x,y)$ $T(x,y,d,t_{i})$ dependence of temperature on depth $(d = depth)$ in timepoint $t_{i}, i \in [1,n]$ , , along a vertical $V_{x,y}$;
Sequence of actions:
1. . Go to «The designer of schemes» and choose scheme type «Well» (see Figure 2).
Figure 2 – Creation of the scheme in vertical direction (Well)
1. Specify a vertical $V_{x,y}$, along which it is necessary to receive the table of dependence of temperatures on depth. For this purpose specify coordinates $(x,y)$ of the vertical of interest in the «Designer of schemes».
2. . Switch the program to the mode of display of temperatures. After switching the scheme will display dependence of temperature on depth, $T(x,y,d,t_{i})$.
3. Open the scheme using double click in the list.
4. . Repeat the following actions for each timepoint $t_{i}$:
• Switch the iteration to timepoint $t_{i}$, for display of value of temperatures at this timepoint;
• Click in a window with the scheme, and press F key to record the scheme (instead of F key it is possible to use the corresponding button on the side control panel (at the left) in a scheme window);
5. All necessary schemes are recorded. Choose the way of export:
• Add the scheme to Microsoft Word (see further How to copy schedules in the report in the Microsoft Word format?) .
• Copy the scheme in Microsoft Excel (see further How to copy schedules in Microsoft Excel?)
Important: It is possible to create schemes of dependence of unfrozen water content on depth, filtration speed on depth, etc. in the same way.
[свернуть]
20. Q: How to copy schemes in the report in the Microsoft Word format? Is it possible to adjust an order of values in the table?
A:
Starting with 3.0.*.699 version all recorded schemes can be added to the report (Microsoft Word format). For export of schemes:
1. Open a scheme window;
2. Press «Add to the Report» button on the side panel at the left (see Figure 1).
3. After the emergence of«Reports Editor» window, set parameters necessary to you:
• «Heading», «Heading of arguments/function values», «Shift for an Argument»;
• Check the box «To remove values with the fixed argument step» and specify an argument step*(see. Figure 1).
4. Press «Close» button. Data of schemes will be copied in the report taking into account the parameters set by you (see Figure 3).
* without the specification of value the table in which the amount of values of an argument will be in proportion to quantity of cells in the computational mesh on axis Z will be added to the report.
Important:This word-processor imposes a number of restrictions for functionality of the user. We recommend to carry out export of schemes to Microsoft Excel that will give the chance to change orientation of the table and to replace the direction of a round.
Figure 1 – Adding schemes to the report.
Figure 2 – Report editor.
Figure 3 – Example of scheme export.
[свернуть]
21. Q: How to copy schemes in Microsoft Excel? Is it possible to change an order of values in the corresponding table automatically?
A: Starting with version 3.0.*.699 it is possible not only to add schemes to the report, but also to copy in Microsoft Excel. Also there is an opportunity to set orientation of the table and the direction of a round automatically:
1. Open a scheme window (see Figure 1);
2. Press «Copy» button on the side panel at the left.
3. After emergence of «Settings of table export» window, set the parameters of copying of the scheme necessary to you.
4. Press «Apply» button. Data of schemes will be copied in a clipboard taking into account the parameters set by you.
5. Open Microsoft Excel document and insert clipboard contents.
Figure 1 – Scheme copy. Settings of table export.
Settings of table export:
1) Using of argument shift. Allows to reckon an argument from the chosen value. We recommend to use it for dependence on depth in case there is an absolute mark concerning which the values of depths are necessary.
2) The fixed argument step. Allows to set an argument step. Values will be interpolated.
3) To replace orientation of the table. Allows to choose horizontal orientation, instead of vertical orientation of the table.
4) To replace the direction of a round of an argument. Allows to display arguments from bigger to smaller.
[свернуть]
22. Q: Is it possible to create the report on entrance data automatically (parameters which the user has set during creation of the project)?
A: Yes, the software provides function of automatic creation of the report on entrance data in Microsoft Excel format.
The call of this function is carried out in a context menu of «Choice of the displayed computation» window.
Figure 1 – Choice of the displayed computation window: сontext menu.
[свернуть]
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2020-02-26 15:24:46
|
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|
https://mathspace.co/textbooks/syllabuses/Syllabus-99/topics/Topic-4539/subtopics/Subtopic-17747/
|
Hong Kong
Stage 4 - Stage 5
# Transformation and Quadratic Graphs
Lesson
As you can see the St. Louis Arch is an example of a parabola in real life. This parabola would be a negative parabola because it is facing down. Parabolas are used very often but people don’t tend to notice. Another example would be the McDonald’s golden arches.
But every parabola is different. So how do we know the equation of the parabola we want?
This lesson focuses on the connection between the equations of quadratics and their graphs, and how we can use the idea of transformations to more easily identify and make sense of these quadratics.
## Equations of Parabolas
The word parabola may not be a common word in everyday life, but the concept of parabolas is most definitely not uncommon. From bridges to radio telescopes, they are not hard to spot. In fact, many famous landmarks are parabolas e.g. the Sydney Harbour Bridge. Parabolas also appear as the trajectory that objects follow once thrown. For example, when you try to make a basketball shot it follows the path of a parabola.
Given how common parabolas are, it would be useful to graph them and know their equations. For example, when avalanche control experts want to trigger an avalanche, they need to know where the detonators will land on the mountain side. If the detonator follows a parabolic path, its equation can help determine such unknowns.
All parabolas can be thought of as some transformation of the standard parabola which has the equation $y=x^2$y=x2.
What would be the equation of a parabola that is flatter or narrower and which has a vertex that is not at $\left(0,0\right)$(0,0)
### Concavity
Concavity is a word used to describe the shape of a curve.
Curves can be concave up (making a cup like shape, or a smiling mouth,)
Curves can be concave down (making a hill shape or a sad mouth, )
$y=x^2$y=x2
• Notice that as the $x$x values we substitute get larger, i.e. $x=2$x=2, $x=5$x=5, $x=50$x=50 etc, $y$y increases towards infinity.
• Notice that as the $x$x values we substitute get negatively larger, i.e. $x=-2$x=2, $x=-5$x=5, $x=-50$x=50 etc, $y$y still increases towards infinity.
• $y$y can never have a negative value.
• This parabola is concave up.
$y=-x^2$y=x2
• Notice that as the $x$x values we substitute get larger, i.e. $x=2$x=2, $x=5$x=5, $x=50$x=50 etc, $y$y decreases towards negative infinity.
• Notice that as the $x$x values we substitute get negatively larger, i.e. $x=-2$x=2, $x=-5$x=5, $x=-50$x=50 etc, $y$y still decreases towards negative infinity.
• $y$y can never have a positive value.
• This parabola is concave down.
The concavity of a parabola is determined by the $a$a value in the equation $y=ax^2$y=ax2.
• $y=-x^2$y=x2 vs $y=x^2$y=x2: The sign of the coefficient $a$a determines whether the parabola is concave up or concave down.
### Narrow or wide
Compare the parabolas $y=\frac{1}{2}x^2$y=12x2$y=x^2$y=x2, and $y=2x^2$y=2x2. The larger the numerical value of $a$a, the more quickly $y$y increases, i.e. the narrower the parabola. In this case $y=2x^2$y=2x2 will be the narrowest.
So just by looking at the coefficient of $x^2$x2, we can determine some features of the parabola's shape:
1. $y=\frac{1}{2}x^2$y=12x2: concave up, wider than $y=x^2$y=x2.
2. $y=-x^2$y=x2: concave down, same narrowness as $y=x^2$y=x2.
3. $y=-3x^2$y=3x2: concave down, narrower than $y=-x^2$y=x2, the result of reflecting $y=3x^2$y=3x2 about the $x$x-axis.
### Horizontal and Vertical translation
Every parabola has a vertex. If it is concave up, the vertex is the point where the $y$y value is a minimum. If it is concave down, the vertex is the point where the $y$y value is a maximum. Keeping this in mind, let's consider the minimum and maximum value of each parabola below to locate its vertex.
##### Vertical Translation
1) $y=x^2$y=x2 (concave up so $y$y has a minimum value at vertex)
• Minimum value of $y$y: $y$y is the result of squaring a value so the smallest possible value is $y=0$y=0
• $x$x value corresponding to minimum $y$y value: we would need to substitute $x=0$x=0 to get $y=0$y=0
• Vertex: $\left(0,0\right)$(0,0)
2) $y=x^2+2$y=x2+2 (same concavity as $y=x^2$y=x2)
• Minimum value of $y$y$y=x^2$y=x2 has a minimum value of $0$0, so adding $2$2 to it, the smallest possible value is $y=2$y=2
• $x$x value corresponding to minimum $y$y value: we would need to substitute $x=0$x=0 to get $y=2$y=2
• Vertex: $\left(0,2\right)$(0,2)
$y=x^2+2$y=x2+2 is the result of a translation of $y=x^2$y=x2 by $2$2 units upwards.
3) $y=-x^2-5$y=x25 (same concavity as $y=-x^2$y=x2)
• Maximum value of $y$y$y=-x^2$y=x2 has a maximum value of $0$0, so subtracting $5$5 from it, the maximum value of $y$y here is $y=-5$y=5
• $x$x value corresponding to minimum $y$y value: we would need to substitute $x=0$x=0 to get $y=-5$y=5
• Vertex: $\left(0,-5\right)$(0,5)
$y=-x^2-5$y=x25 is the result of a translation of $y=-x^2$y=x2 by $5$5 units downwards.
Considering vertical translations only, the equations we encounter will be of the form $y=ax^2+k$y=ax2+k, where:
• $a$a determines concavity (positive or negative, narrow or wide)
• $k$k represents the vertical shift upwards (for $k>0$k>0) or downwards (for $k<0$k<0)
##### Horizontal Translation
4) $y=\left(x-3\right)^2$y=(x3)2 (concave up so $y$y has a minimum value at vertex)
• Minimum value of $y$y: $y$y is still the result of squaring a value so the smallest possible value is $y=0$y=0
• $x$x value corresponding to minimum $y$y value: we would need to substitute $x=3$x=3 to get $y=0$y=0
• Vertex: $\left(3,0\right)$(3,0)
$y=\left(x-3\right)^2$y=(x3)2 is the result of a $3$3 unit translation of $y=x^2$y=x2 to the right.
5) $y=\left(x+3\right)^2$y=(x+3)2 (concave up so $y$y has a minimum value at vertex)
• Minimum value of $y$y: $y$y is still the result of squaring a value so the smallest possible value is $y=0$y=0
• $x$x value corresponding to minimum $y$y value: here we would need to substitute $x=-3$x=3 to get $y=0$y=0
• Vertex: $\left(-3,0\right)$(3,0)
$y=\left(x+3\right)^2$y=(x+3)2 is the result of $3$3 unit translation of $y=x^2$y=x2 to the left.
6) $y=-\left(x-4\right)^2$y=(x4)2 (concave down so $y$y has a maximum value at vertex)
• Maximum value of $y$y: $y$y is the result of squaring a value, and then multiplying that square value by $-1$1, so the largest possible value is $y=0$y=0
• $x$x value corresponding to maximum $y$y value: here we would need to substitute $x=4$x=4 to get $y=0$y=0
• Vertex: $\left(4,0\right)$(4,0)
$y=-\left(x-4\right)^2$y=(x4)2 is the result of a $4$4 unit translation of $y=x^2$y=x2 to the right.
Now that we know how to represent vertical and horizontal translations algebraically, we can summarise the key concepts in the form $y=a\left(x-h\right)^2+k$y=a(xh)2+k, where:
• $a$a determines concavity
• $k$k represents the vertical shift upwards or downwards
• $h$h represents the horizontal shift to the right or to the left
In this form, every parabola can be thought of as some sort of transformation of $y=x^2$y=x2. All we need to consider by looking at the equation is:
• Is there any narrowing, widening or reflecting of the curve $y=x^2$y=x2?
• Is there any horizontal or vertical translation of the curve $y=x^2$y=x2?
#### Practice questions
##### Question 1
A parabola of the form $y=ax^2$y=ax2 goes through the point $\left(2,-4\right)$(2,4).
1. What is the value of $a$a?
2. What are the coordinates of the vertex?
Vertex $=$=$\left(\editable{},\editable{}\right)$(,)
3. Plot the graph of the parabola.
##### Question 2
Consider the parabola $y=8+3\left(x-7\right)^2$y=8+3(x7)2.
1. What is the vertex of the parabola? Give your answer in the form $\left(h,k\right)$(h,k).
2. The parabola is reflected about the $y$y-axis. What will its new equation be?
3. What will be the vertex of the new parabola formed after the reflection?
##### Question 3
1. Write down the equation of the new parabola when $y=x^2$y=x2 is translated $6$6 units to the right and $2$2 units down.
2. Graph the parabola formed after the translations.
##### Question 4
A skydiving instructor wants to use the equation $y=at^2+c$y=at2+c to model the height in metres of a skydiver above the ground $t$t seconds after jumping out of the plane.
1. Which of the following would be the correct values of $a$a and $c$c to use?
Negative $a$a, positive $c$c
A
Positive $a$a, negative $c$c
B
Negative $a$a, negative $c$c
C
Positive $a$a, positive $c$c
D
2. She does a test run, jumping out of the plane at a height of $2560$2560 metres. Find the value of $c$c.
3. After $160$160 seconds, the instructor lands on the ground. Form an equation to find the value of $a$a.
4. After $23$23 seconds, how far above the ground was she?
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2023-03-26 03:32:04
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|
https://proofwiki.org/wiki/Characteristic_Function_of_Intersection/Variant_1
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# Characteristic Function of Intersection/Variant 1
## Theorem
Let $A, B \subseteq S$.
Let $\chi_{A \cap B}$ be the characteristic function of their intersection $A \cap B$.
Then:
$\chi_{A \cap B} = \chi_A \chi_B$
## Proof
By Characteristic Function Determined by 1-Fiber, it suffices to show that:
$\map {\chi_A} s \, \map {\chi_B} s = 1 \iff s \in A \cap B$
Now, both $\chi_A$ and $\chi_B$ are characteristic functions.
It follows that, for any $s \in S$:
$\map {\chi_A} s \, \map {\chi_B} s = 1 \iff \map {\chi_A} s = \map {\chi_B} s = 1$
By definition of $\chi_A$ and $\chi_B$, this is equivalent to the statement that both $s \in A$ and $s \in B$.
That is, $s \in A \cap B$, by definition of set intersection.
$\blacksquare$
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2020-07-14 15:40:01
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|
https://gharpedia.com/remove-dents-wooden-surface/
|
## Steps to Remove Dents from the Wooden Surface
Wooden surfaces are damaged due to the dropping of heavy tools or object on it. It also gets damaged due to movement of furniture. Small and shallow dents in soft woods are usually easy to remove. Large and deep dents in hard wood are hard to repair.
Courtesy - Flickr
Follow the below-given steps to remove dents from the wooden surface:
• Wet the affected area by using warm water. Not too much, only enough to cover the dents.
• After that cover the dents with a paper towel. Set the paper towel proper, so that it does not fold over.
• Set the iron to high steam and hold it over the affected area and make small movements back and forth in a circle. Press the paper towel when it dries. It won’t take a long time to evaporate.
• At this time, the wood fibres that have absorbed the water will expand so as to come back to where they originally were.
• This process will be continued and repeated by adding more water until the dents rise to get flushed with the rest of the material.
• However, the area around dent is pretty smooth. Outline of the affected area is seen after this process. Remove this outline by rubbing with sandpaper. After that apply tung oil. Even after all this process, the finish may probably not the same as the original one, but the area will definitely look better than before.
It is the simplest method without spending money to remove the dents from the soft wooden surface. Large and deeper cracks in the wooden surface may require additional work under similar this process.
## Material Exhibition
Explore the world of materials.
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2019-03-23 07:28:00
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|
https://mindmatters.ai/t/brian-thomas/
|
Mind Matters Natural and Artificial Intelligence News and Analysis
# TagBrian Thomas
## Haiti: Turning the Blazing Sun Into a Power Source
Brian Thomas and Kayla Garrett learned how to make solar power work in Haiti by listening to people, especially the Haitians they work with
In podcast Episode 209, Robert J. Marks continues the discussion with Brian Thomas and Kayla Garrett of JustEnergy about appropriate technology for energy-starved Haiti: Solar powering hospitals, orphanages & schools (October 20, 2022): Robert J. Marks: I was informed that people [in Haiti] on average make a dollar a day and they have to go out and they have to buy gas sometimes on the black market for $20,$30 a gallon. It’s just crazy.. So one of the things that you’re concentrated on as engineers is to increase the energy access to Haitians. So what’s the technology that you use to increase the energy access? Brian Thomas: Haiti doesn’t have any petroleum — any oil, gasoline, diesel… or even Read More ›
## Appropriate Technology: Solar Powering Hospitals, Orphanages & Schools
Under normal conditions only about 30% of the entire country of Haiti has physical connection or access to the electricity grid. Electrical engineer Brian Thomas and environmental engineer Kayla Garrett work for a small nonprofit called JustEnergy, which works in northern Haiti creating solar energy systems to power hospitals, clinics, schools, and orphanages to combat this lack of access to Read More ›
## How Solar Energy Ran a Haitian Hospital During the Energy War
Gangs seized control of the ports at which ships bringing fuel docked, cutting off supplies, in an effort to force the Acting President to step down
Yesterday, we looked at the first part of the “Appropriate Technology: the Haitian Energy Problem” podcast (October 13, 2022). Walter Bradley Center director Robert J. Marks interviewed engineers Brian Thomas and Kayla Garrett of JustEnergy on the current shortage of energy sources in Haiti. In the second part of the podcast, they look at what might be done: https://mindmatters.ai/wp-content/uploads/sites/2/2022/10/Mind-Matters-208-Brian-Thomas-Kayla-Garrett.mp3 A partial transcript, notes, and additional resources follow. Brian Thomas: Let’s stop and think. If gasoline is $20,$25 a gallon — even if it’s \$10 a gallon — and you make very little money or you don’t have a job at all… tThen gasoline is like cash. You can sell that. You can turn around and sell that. So gasoline Read More ›
## In Haiti, Debates Over Electric vs. Gas-Powered Cars Are a Luxury
Never mind self-driving cars. The quest for “just enough” energy is a daily, sometimes life-and-death issue, as Kayla Garrett and Brian Thomas tell Robert J. Marks
In “Appropriate Technology: the Haitian Energy Problem” (October 13, 2022), Walter Bradley Center director Robert J. Marks interviewed engineers Brian Thomas and Kayla Garrett on a critical question: meeting the energy needs of a developing nation like Haiti sustainable — the only way it can be done: https://mindmatters.ai/wp-content/uploads/sites/2/2022/10/Mind-Matters-208-Brian-Thomas-Kayla-Garrett.mp3 A partial transcript, notes, and additional resources follow. Robert J. Marks: Not all countries need the latest technologies. Those in Third World countries don’t need high powered computers or the latest car from Tesla. They have more fundamental concerns like, how do I feed my family tomorrow? Where do I get clean water? And where can I get power? These needs typically do not involve the latest edge cutting technology. Supplying needed Read More ›
## Appropriate Technology: The Haitian Energy Problem
Not all countries need the latest technologies. Those in developing countries don’t need high powered computers or the latest car from Tesla. They have more fundamental concerns like how do I feed my family tomorrow? Where do I get clean water? And where can I get power and energy? This effort of supplying needed technology is referred to as appropriate Read More ›
## Walter Bradley: Finding a Life of Greater Purpose
Bradley has been a pioneer in the development of appropriate technologies for developing regions of the world
In last week’s podcast, “The Life of Walter Bradley With William Dembski (Part I),” Walter Bradley Center director Robert J. Marks and design theorist William Dembski discuss the biography they have written about a remarkable engineer, Walter Bradley, For a Greater Purpose: The Life and Legacy of Walter Bradley. It also helps explain why we call ourselves the Walter Bradley Center, as we seek to extend Dr. Bradley’s work. https://episodes.castos.com/mindmatters/Mind-Matters-121-William-Dembski.mp3 A partial transcript follows. This transcript begins at 02:55. Show notes and links follow. Before getting down to the main business, design theorist William Dembski, possibly the best known theorist of design in nature, told Robert Marks that he plans a second edition of his Cambridge University Press book, The Read More ›
## Technology Only Works If People Use It
Introducing appropriate technology to an old culture includes careful listening, says engineering prof Brian Thomas
"We cannot assume that because it works in our culture, it will work in another culture," he says.
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2022-12-03 03:11:04
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https://datascienceplus.com/building-barplots-with-error-bars/
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An online community for showcasing R & Python tutorials. It operates as a networking platform for data scientists to promote their talent and get hired. Our mission is to empower data scientists by bridging the gap between talent and opportunity.
Visualizing Data
# Building Barplots with Error Bars
Bar charts are a pretty common way to represent data visually, but constructing them isn’t always the most intuitive thing in the world.
One way that we can construct these graphs is using R’s default packages.
## Barplots using base R
Let’s start by viewing our dataframe: here we will be finding the mean miles per gallon by number of cylinders and number of gears.
View(mtcars)
We begin by aggregating our data by cylinders and gears and specify that we want to return the mean, standard deviation, and number of observations for each group:
myData <- aggregate(mtcars$mpg, by = list(cyl = mtcars$cyl, gears = mtcars$gear), FUN = function(x) c(mean = mean(x), sd = sd(x), n = length(x))) After this, we’ll need to do a little manipulation since the previous function returned matrices instead of vectors myData <- do.call(data.frame, myData) And now let's compute the standard error for each group. We can then rename the columns just for ease of use. myData$se <- myData$x.sd / sqrt(myData$x.n)
colnames(myData) <- c("cyl", "gears", "mean", "sd", "n", "se")
myData$names <- c(paste(myData$cyl, "cyl /",
myData$gears, " gear")) Now we’re in good shape to start constructing our plot! Here, we’ll start by widening the plot margins just a tad so that nothing runs off the edge of the figure (using the par() function). It’s also a good habit to specify the upper bounds of your plot since the error bars are going to extend past the height of your bars. Beyond this, it’s just any additional aesthetic styling that you want to tweak and you’re good to go! The error bars are added in at the end using the segments() and arrows() functions. In this case, we are extending the error bars to ±2 standard errors about the mean. par(mar = c(5, 6, 4, 5) + 0.1) plotTop <- max(myData$mean) +
myData[myData$mean == max(myData$mean), 6] * 3
barCenters <- barplot(height = myData$mean, names.arg = myData$names,
beside = true, las = 2,
ylim = c(0, plotTop),
cex.names = 0.75, xaxt = "n",
main = "Mileage by No. Cylinders and No. Gears",
ylab = "Miles per Gallon",
border = "black", axes = TRUE)
# Specify the groupings. We use srt = 45 for a
# 45 degree string rotation
text(x = barCenters, y = par("usr")[3] - 1, srt = 45,
adj = 1, labels = myData$names, xpd = TRUE) segments(barCenters, myData$mean - myData$se * 2, barCenters, myData$mean + myData$se * 2, lwd = 1.5) arrows(barCenters, myData$mean - myData$se * 2, barCenters, myData$mean + myData$se * 2, lwd = 1.5, angle = 90, code = 3, length = 0.05) This will give us a barplot that looks like this: ### Grouped barplots But… that’s kind of an ugly graph. Wouldn’t it be nicer if we could group the bars by number of cylinders or number of gears? Turns out, R makes this pretty easy with just a couple of tweaks to our code! Instead of columns of means, we just need to supply barplot() with a matrix of means. I.e., instead of this: head(myData) cyl gears mean sd n se names 4 3 21.500 NA 1 NA 4 cyl / 3 gear 4 4 26.925 4.807360 8 1.6996586 4 cyl / 4 gear 4 5 28.200 3.111270 2 2.2000000 4 cyl / 5 gear 6 3 19.750 2.333452 2 1.6500000 6 cyl / 3 gear 6 4 19.750 1.552417 4 0.7762087 6 cyl / 4 gear 6 5 19.700 NA 1 NA 6 cyl / 5 gear we supply: tapply(myData$mean, list(myData$cyl, myData$gears),
function(x) c(x = x))
3 4 5
4 21.50 26.925 28.2
6 19.75 19.750 19.7
8 15.05 NA 15.4
All that this requires is that we switch out a couple arguments in our previous code, resulting in:
tabbedMeans <- tapply(myData$mean, list(myData$cyl,
myData$gears), function(x) c(x = x)) tabbedSE <- tapply(myData$se, list(myData$cyl, myData$gears),
function(x) c(x = x))
barCenters <- barplot(height = tabbedMeans,
beside = TRUE, las = 1,
ylim = c(0, plotTop),
cex.names = 0.75,
main = "Mileage by No. Cylinders and No. Gears",
ylab = "Miles per Gallon",
xlab = "No. Gears",
border = "black", axes = TRUE,
legend.text = TRUE,
args.legend = list(title = "No. Cylinders",
x = "topright",
cex = .7))
segments(barCenters, tabbedMeans - tabbedSE * 2, barCenters,
tabbedMeans + tabbedSE * 2, lwd = 1.5)
arrows(barCenters, tabbedMeans - tabbedSE * 2, barCenters,
tabbedMeans + tabbedSE * 2, lwd = 1.5, angle = 90,
code = 3, length = 0.05)
This, in turn, gives us a nicely grouped barplot:
## Barplots using ggplot2
Unfortunately, that’s a really messy solution. It’s a lot of code written for a relatively small return. There’s got to be an easier way to do this, right?
Thankfully, there is! Alternately, we can use Hadley Wickham’s ggplot2 package to streamline everything a little bit. We’ll use the myData data frame created at the start of the tutorial. After loading the library, everything follows similar steps to what we did above. Here we start by specifying the dodge (the spacing between bars) as well as the upper and lower limits of the x and y axes.
After this, we construct a ggplot object that contains information about the data frame we’re using as well as the x and y variables. From there it’s a simple matter of plotting our data as a barplot (geom_bar()) with error bars (geom_errorbar())!
library(ggplot2)
dodge <- position_dodge(width = 0.9)
limits <- aes(ymax = myData$mean + myData$se,
ymin = myData$mean - myData$se)
p <- ggplot(data = myData, aes(x = names, y = mean, fill = names))
p + geom_bar(stat = "identity", position = dodge) +
geom_errorbar(limits, position = dodge, width = 0.25) +
theme(axis.text.x=element_blank(), axis.ticks.x=element_blank(),
axis.title.x=element_blank())
This results in a similar barplot as before:
### Grouped barplots
Just as before, we can also group our bars. Let’s try grouping by number of cylinders this time:
limits <- aes(ymax = myData$mean + myData$se,
ymin = myData$mean - myData$se)
p <- ggplot(data = myData, aes(x = factor(cyl), y = mean,
fill = factor(gears)))
p + geom_bar(stat = "identity",
position = position_dodge(0.9)) +
geom_errorbar(limits, position = position_dodge(0.9),
width = 0.25) +
labs(x = "No. Cylinders", y = "Miles Per Gallon") +
ggtitle("Mileage by No. Cylinders\nand No. Gears") +
scale_fill_discrete(name = "No. Gears")
In all cases, you can fine-tune the aesthetics (colors, spacing, etc.) to your liking. For example, by fiddling with some colors and font sizes:
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2017-10-20 06:59:20
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https://math.stackexchange.com/questions/750835/definite-integral-in-spherical-coordinates
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# Definite integral in spherical coordinates
I want to compute the volume enclosed by the sphere $x^2+y^2+z^2\le4$ with $0\le z\le 1.$
If I use the rings method I get:
$$\pi \int_0^1 \left( \sqrt{4-z^2}\right)^2dz=\frac{11}{3}\pi$$
If I set up a triple integral in cylindrical coordinates I get:
$$\int_0^{2\pi} \int_0^1 \int_0^{\sqrt{4-z^2}} r\,dr\,dz\,d\theta=\frac{11}{3} \pi$$
So far so good, but I gan't get it done with a triple integral in spherical coordinates. I assume the second integral is wrong but I don't know why:
$$\int_0^{2\pi} \int_0^{\pi/6} \int_0^{2} \rho^2 \sin(\phi)\,d\rho\,d\phi\,d\theta=\left( \int_0^{2\pi}d\theta \right)\left( \int_0^{\pi/6}\sin(\phi) \,d\phi\right)\left( \int_0^{2} \rho^2 \,d\rho \right)\neq \frac{11}{3} \pi$$
You should split the triple integral into two parts. $$V = \int_0^{2\pi} \int_0^{\frac{\pi}{3}} \int_0^{\frac{1}{cos \phi}} \rho^2 \sin(\phi) \,d\rho \,d\phi \,d\theta+ \int_0^{2\pi} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \int_0^2 \rho^2 \sin(\phi) \,d\rho \,d\phi \,d\theta$$ Note that the first one is the volume of a cone.
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2019-09-19 15:19:33
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https://www.physicsforums.com/threads/probability-question.81223/
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# Homework Help: Probability question
1. Jul 5, 2005
### Townsend
If event S has a 40 percent chance of happening on the first try and a 65 percent chance of happening on the second try what is the chance of event A happening?
Let A be the event S happens on the first try, note $$A^c$$ is the complement of A.
Let B be the event S happens on the second try.
I reasoned it to be a 79 percent chance of S happening.
1. A and B are mutually exclusive, in other words if A happens B does not, vise versa.
2. The probability of event B happening depends on event A not happening.
Thus $$P(B \cap A^c) = .39$$, so P(S) = P(A) + $$P(B \cap A^c)$$ = .79.
So I reason a 79 percent chance of event S.
Thanks
2. Jul 5, 2005
### honestrosewater
You mean of event S happening?
Why do you think they are mutually exclusive? Does the problem say so? Why do you think they are dependent?
The first paragraph asks for the probability that, in two tries, S occurs. That would mean either S occurs on the first try or S occurs on the second try or S occurs on both tries. Think of getting heads on two coin tosses. Either the first toss is heads or the second toss is heads or both tosses are heads. What formula do you use for these "or" situations?
3. Jul 5, 2005
### Townsend
Yes, sorry.
I didn't state the problem as it was stated in the text to save some typing. The problem is
A salesman has a 40 percent chance of making a sale the first time he visits a house and a 65 percent chance of making a sale the second time he visits that same house and he will only ever visit a house twice. What is the chance he will make a sale?
Based on this I can only reason that he will either make a sale the first day or second day but not both days. So I suppose that makes the two events mutually exclusive.
4. Jul 5, 2005
### honestrosewater
Yeah, that's not the greatest. Maybe he's selling different things each time, there could be different people living in the house, etc. Go with whatever you're willing to defend. I would go with the most straightforward- A and B are non-mutually exclusive, independent events. You could also solve it for each scenario. Do you have the formula for P(A or B) (or P(A $\cup$ B))?
Last edited: Jul 5, 2005
5. Jul 5, 2005
### Townsend
Not sure... If A is an event and B is an event then the sample points in A or B would just be A union B, right? The problem is figuring out what those sample points are. If you accept that B is dependent on A and A intersect B is null then how do you find P(A or B) if you're only given what P(A) and P(B) are, and P(B) is given with the assumption that A will not happen?
6. Jul 5, 2005
### honestrosewater
Right, P(A or B) is the same as P(A U B). P(A U B) = P(A) + P(B) - P(A $\cap$ B).
If A and B are mutually exclusive, what is P(A $\cap$ B)?
If A and B are independent, what is P(A $\cap$ B)?
So why would you think A and B are dependent (but not mutually exclusive)? In your original post, you said you thought A and B were mutually exclusive. If they are, the probability of A occurring given that B has occurred is what? The probability of B occurring given that A has occurred is what?
The reason I would go with A and B being independent is that, by default, you don't assume more than the problem tells you. If A and B were mutually exclusive, the problem should have added:
"But if he makes a sale the first try, he doesn't make a sale the second try, and if he makes a sale the second try...." or something similar.
If A and B were dependent, the problem should have added:
"If he makes a sale the first try, then on the second try..."
or something else to tell you how the events depend on each other. It doesn't say anything about the events being dependent or mutually exclusive, and there's nothing inherent in the situation that makes the events dependent or mutually exclusive, so you should just assume the events are independent. You could argue that a salesman wouldn't even try to make a sale on the second try if he made a sale on the first try, but you should be prepared to really make that argument.
Last edited: Jul 5, 2005
7. Jul 5, 2005
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2018-06-19 07:23:28
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https://www.aimsciences.org/article/doi/10.3934/eect.2022003
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Article Contents
Article Contents
# A shape optimization problem constrained with the Stokes equations to address maximization of vortices
• *Corresponding author: John Sebastian Simon
• We study an optimization problem that aims to determine the shape of an obstacle that is submerged in a fluid governed by the Stokes equations. The mentioned flow takes place in a channel, which motivated the imposition of a Poiseuille-like input function on one end and a do-nothing boundary condition on the other. The maximization of the vorticity is addressed by the $L^2$-norm of the curl and the det-grad measure of the fluid. We impose a Tikhonov regularization in the form of a perimeter functional and a volume constraint to address the possibility of topological change. Having been able to establish the existence of an optimal shape, the first order necessary condition was formulated by utilizing the so-called rearrangement method. Finally, numerical examples are presented by utilizing a finite element method on the governing states, and a gradient descent method for the deformation of the domain. On the said gradient descent method, we use two approaches to address the volume constraint: one is by utilizing the augmented Lagrangian method; and the other one is by utilizing a class of divergence-free deformation fields.
Mathematics Subject Classification: Primary: 49Q10, 49J20, 49K20; Secondary: 35Q93.
Citation:
• Figure 1. Set up of the domain
Figure 2. Initial geometry of the domain with the refined mesh
Figure 3. From curlaL-problem, the figure features the following: (left) Evolution of the free-boundary $\Gamma_{\rm f}$, (upper right) Normalized trend of the objective functional, (lower right) Normalized trend of the volume
Figure 4. From detgradaL-problem, the figure features the following: (left) Evolution of the free-boundary $\Gamma_{\rm f}$, (upper right) Normalized trend of the objective functional, (lower right) Normalized trend of the volume
Figure 5. From curldF-problem, the figure features the following: (left) Evolution of the free-boundary $\Gamma_{\rm f}$, (upper right) Normalized trend of the objective functional, (lower right) Normalized trend of the volume
Figure 6. (left) Comparison of final shapes between aL-algorithm and dF-algorithm for the problem with the parameters $\gamma_1 = 1$, $\gamma_2 = 0$ and $\alpha = 5$, (upper right) Comparison of objective value trends between aL-algorithm and dF-algorithm, (lower right) Comparison of volume trends between aL-algorithm and dF-algorithm
Figure 7. From detgraddF-problem, the figure features the following: (left) Evolution of the free-boundary $\Gamma_{\rm f}$, (upper right) Normalized trend of the objective functional, (lower right) Normalized trend of the volume
Figure 8. (left) Comparison of final shapes between aL-algorithm and dF-algorithm with the parameters $\gamma_1 = 0$, $\gamma_2 = 1$, and $\alpha = 1$, (upper right) Comparison of objective value trends between aL-algorithm and dF-algorithm, (lower right) Comparison of volume trends between aL-algorithm and dF-algorithm
Figure 9. (left) Comparison of final shapes between aL-algorithm and dF-algorithm with the parameters $\gamma_1 = 0$, $\gamma_2 = 1$, and $\alpha = 1.2$, (upper right) Comparison of objective value trends between aL-algorithm and dF-algorithm, (lower right) Comparison of volume trends between aL-algorithm and dF-algorithm
Figure 10. From configuration 1 of mixeddF-problem, the figure features the following: (left) Evolution of the free-boundary $\Gamma_{\rm f}$, (upper right) Normalized trend of the objective functional, (lower right) Normalized trend of the volume
Figure 11. (left) Plots of the solutions of mixeddF-problem (using configuration 1), of curldF-problem and of detgraddF-problem generated boundaries, (upper right) Comparison of the objective function value of the curldf-problem and the curl part of the mixeddF-problem, (lower right) Comparison of the objective function value of the curldF-problem and the detgrad part of the mixeddf-problem
Figure 12. (left) Plots of the solutions of mixeddF-problem (using configuration 2), of curldF-problem and of detgraddF-problem generated boundaries, (upper right) Comparison of the objective function value of the curldF-problem and the detgrad part of the mixeddF-problem, (lower right) Comparison of the objective function value of the curldF-problem and the detgrad part of the mixeddF-problem
Figure 13. (left) Plots of the solutions of mixeddF-problem (using configuration 4), of curldF-problem and of detgraddF-problem generated boundaries, (upper right) Comparison of the objective function value of the curldF-problem and the detgrad part of the mixeddF-problem, (lower right) Comparison of the objective function value of the curldF-problem and the detgrad part of the mixeddF-problem
Figure 14. (left) Plots of the solutions of mixeddF-problem (configurations 1 and 10), of curldF-problem, and of detgraddF-problem, (upper right) Hausdorff distance between mixeddF-solution and curldF-solution on each configuration, (lower right) Hausdorff distance between mixeddF-solution and detgraddF-solution on each configuration
Figure 15. The figure shows the configurations for the two-obstacle experiment: (top) Set-up of the domain where the obstacles are placed parallel to the flow; (bottom) Set-up of the domain where the obstacles are placed perpendicular to the flow
Figure 16. From the first configuration of the two-obstacle experiment, the figure features the following: (left) Evolution of the free-boundary $\Gamma_{\rm f}$, (upper right) Normalized trend of the objective functional, (lower right) Normalized trend of the volume
Figure 17. From the second configuration of the two-obstacle experiment, the figure features the following: (left) Evolution of the free-boundary $\Gamma_{\rm f}$, (upper right) Normalized trend of the objective functional, (lower right) Normalized trend of the volume
Figure 18. The figure shows the comparison of the flows using the initial domain (lower part), and the final shape from curldf-problem (upper part)
Figure 19. The figure shows the comparison of the flows using the initial domain (lower part), and the final shape from detgraddf-problem (upper part)
Figure 20. The figure shows the comparison of the flows using the the final shape from curldf-problem (lower part), and the final shape from detgraddf-problem (upper part)
Table 1. Parameter values for curlaL-problem
Parameter Value Parameter Value $\alpha$ 6.0 $\ell_0$ 20 $b_0$ $1\times10^{-4}$ $\tau$ 1.05 $\overline{b}$ 10
Table 2. Parameter values for detgradaL-problem
Parameter Value Parameter Value $\alpha$ 1.3 $\ell_0$ .5 $b_0$ $1\times10^{-2}$ $\tau$ 1.05 $\overline{b}$ 10
Table 3. Parameter Values for mixeddF-problem
Configuration $\alpha$ $\gamma_1$ $\gamma_2$ 1 6.0 1.0 1.0 2 7.0 1.0 2.0 3 8.0 1.0 3.0 4 9.0 1.0 4.0 5 10.0 1.0 5.0 6 11.0 1.0 6.0 7 12.0 1.0 7.0 8 13.0 1.0 8.0 9 14.0 1.0 9.0 10 15.0 1.0 10.0
• [1] H. Azegami, Solution of shape optimization problem and its application to product design, In Mathematical Analysis of Continuum Mechanics and Industrial Applications, (eds. H. Itou, M. Kimura, V. Chalupecký, K. Ohtsuka, D. Tagami and A. Takada), Springer Singapore, Singapore, 26 (2017), 83–98. doi: 10.1007/978-981-10-2633-1_6. [2] J. Bacani and G. Peichl, On the first-order shape derivative of the Kohn–Vogelius cost functional of the Bernoulli problem, Abstr. Appl. Anal., 2013 (2013), Art. ID 384320, 19 pp. doi: 10.1155/2013/384320. [3] D. Chenais, On the existence of a solution in a domain identification problem, J. Math. Anal. Appl, 52 (1975), 189-219. doi: 10.1016/0022-247X(75)90091-8. [4] M. S. Chong, A. E. Perry and B. J. Cantwell, A general classification of three-dimensional flow fields, Phys. Fluids A, 2 (1990), 765-777. doi: 10.1063/1.857730. [5] C. Dapogny, P. Frey, F. Omnès and Y. Privat, Geometrical shape optimization in fluid mechanics using {F}ree{F}em++, Struct. Multidiscip. Optim., 58 (2018), 2761-2788. doi: 10.1007/s00158-018-2023-2. [6] M. Delfour and J.-P. Zolesio, Shapes and Geometries: Metrics, Analysis, Differential Calculus, and Optimization, 2$^{nd}$ edition, Society for Industrial and Applied Mathematics (SIAM), Philadelphia, PA, 2011. doi: 10.1137/1.9780898719826. [7] M. Desai and K. Ito, Optimal controls of Navier–Stokes equations, SIAM J. Control Optim., 32 (1994), 1428-1446. doi: 10.1137/S0363012992224972. [8] M. F. Eggl and P. J. Schmid, Mixing enhancement in binary fluids using optimised stirring strategies, J. Fluid Mech., 899 (2020), A24, 21 pp. doi: 10.1017/jfm.2020.448. [9] L. Evans, Partial Differential Equations, Graduate Studies in Mathematics, American Mathematical Society, 1998. [10] T. L. B. Flinois and T. Colonius, Optimal control of circular cylinder wakes using long control horizons, Physics of Fluids, 27 (2015), 087105. doi: 10.1063/1.4928896. [11] Z. M. Gao, Y. C. Ma and H. W. Zhuang, Shape optimization for Navier–Stokes flow, Inverse Probl. Sci. Eng., 16 (2008), 583-616. doi: 10.1080/17415970701743319. [12] V. Girault and P.-A. Raviart, Finite Element Methods for Navier–Stokes Equations: Theory and Algorithms, Springer Series in Computational Mathematics, 5. Springer-Verlag, Berlin, 1986. doi: 10.1007/978-3-642-61623-5. [13] K. Goto, K. Nakajima and H. Notsu, Twin vortex computer in fluid flow, New J. Phys., 23 (2021), 063051, 14 pp. doi: 10.1088/1367-2630/ac024d. [14] J. Haslinger, K. Ito, T. Kozubek, K. Kunisch and G. Peichl, On the shape derivative for problems of Bernoulli type, Interfaces Free Bound., 11 (2009), 317-330. doi: 10.4171/IFB/213. [15] J. Haslinger, J. Málek and J. Stebel, Shape optimization in problems governed by generalised Navier–Stokes equations: Existence analysis, Control Cybernet., 34 (2005), 283-303. [16] F. Hecht, New development in FreeFem++, J. Numer. Math., 20 (2012), 251-265. doi: 10.1515/jnum-2012-0013. [17] A. Henrot and M. Pierre, Shape Variation and Optimization: A Geometrical Analysis, EMS Tracts in Mathematics, 28. European Mathematical Society (EMS), Zürich, 2018. doi: 10.4171/178. [18] A. Henrot and Y. Privat, What is the optimal shape of a pipe?, Arch. Ration. Mech. Anal., 196 (2010), 281-302. doi: 10.1007/s00205-009-0243-8. [19] J. C. R. Hunt, A. A. Wray and P. Moin, Eddies, streams, and convergence zones in turbulent flows, In Proc. 1988 Summer Program of Center for Turbulence Research Program, (1998), 193–208. [20] K. Ito, K. Kunisch and G. H. Peichl, Variational approach to shape derivatives, ESAIM Control Optim. Calc. Var., 14 (2008), 517-539. doi: 10.1051/cocv:2008002. [21] Y. Iwata, H. Azegami, T. Aoyama and E. Katamine, Numerical solution to shape optimization problems for non-stationary Navier–Stokes problems, JSIAM Lett., 2 (2010), 37-40. doi: 10.14495/jsiaml.2.37. [22] J. Jeong and F. Hussain, On the identification of a vortex, J. Fluid Mech., 285 (1995), 69-94. doi: 10.1017/S0022112095000462. [23] H. Kasumba and K. Kunisch, Vortex control in channel flows using translational invariant cost functionals, Comput. Optim. Appl., 52 (2012), 691-717. doi: 10.1007/s10589-011-9434-y. [24] G. Mather, I. Mezić, S. Grivopoulos, U. Vaidya and L. Petzold, Optimal control of mixing in Stokes fluid flows, J. Fluid Mech., 580 (2007), 261-281. doi: 10.1017/S0022112007005332. [25] B. Mohammadi and O. Pironneau, Applied Shape Optimization for Fluids, 2$^{nd}$ edition, Numerical Mathematics and Scientific Computations, Oxford University Press, 2010. [26] P. Monk, Finite Element Methods for Maxwell's Equations, Oxford University Press, 2003. doi: 10.1093/acprof:oso/9780198508885.001.0001. [27] M. T. Nair, M. Hegland and R. S. Anderssen, The trade-off between regularity and stability in Tikhonov regularization, Math. Comp., 66 (1997), 193-206. doi: 10.1090/S0025-5718-97-00811-9. [28] J. Nocedal and S. Wright, Numerical Optimization, vol. 2 of Springer Series in Operations Research and Financial Engineering, 2$^{nd}$ edition, Springer, New York, 2006. [29] H. Notsu and M. Tabata, Error estimates of a stabilized Lagrange-Galerkin scheme for the Navier–Stokes equations, ESAIM Math. Model. Numer. Anal., 50 (2016), 361-380. doi: 10.1051/m2an/2015047. [30] M. Pošta and T. Roubíček, Optimal control of Navier–Stokes equations by Oseen approximation, Comput. Math. Appl., 53 (2007), 569-581. doi: 10.1016/j.camwa.2006.02.034. [31] J. F. T. Rabago and H. Azegami, An improved shape optimization formulation of the Bernoulli problem by tracking the Neumann data, J. Engrg. Math., 117 (2019), 1-29. doi: 10.1007/s10665-019-10005-x. [32] J. F. T. Rabago and H. Azegami, A second-order shape optimization algorithm for solving the exterior Bernoulli free boundary problem using a new boundary cost functional, Comput. Optim. Appl., 77 (2020), 251-305. doi: 10.1007/s10589-020-00199-7. [33] S. Schmidt and V. Schulz, Shape derivatives for general objective functions and the incompressible Navier–Stokes equations, Control Cybernet., 39 (2010), 677-713. [34] J. Sokolowski and J.-P. Zolesio, Introduction to Shape Optimization: Shape Sensitivity Analysis, 1$^{st}$ edition, Springer Series in Computational Mathematics, 16. Springer-Verlag Berlin, 1992. doi: 10.1007/978-3-642-58106-9.
Open Access Under a Creative Commons license
Figures(20)
Tables(3)
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2023-03-30 07:55:02
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https://space.stackexchange.com/questions/33331/how-to-reach-a-target-apoapsis-with-a-gravity-turn
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# How to reach a target apoapsis with a gravity turn?
Ok, here I go.
I have as a project to simulate the launch of a payload by a rocket into polar orbit. Since original question was too big, let's get it into separate parts. First of all, trajectory and movement.
Considering the only experience we ever got with orbital mechanics and rocket launches was KSP, I am going to base my program on it.
So, as far as I understand, to launch something into a circular orbit, you need to perform a gravity turn. KSP seems to be doing it by increasing the pitch angle over time and when your periapsis is at the desired height, your pitch angle should change to 90 degrees(or 0-perpendicular to vertical anyways) and thrust in order to get the required speed. From this I tried to come up with a rough algorithm in matlab which plots the trajectory up to 600km, where it reaches with an x-speed(horizontal) of 7.8km/s(which seems fit for polar orbit) and -1.2m/s verical speed based on trial and error method. But I believe I can make this more efficient if I implement the KSP method, aka calculate the periapsis at each step until it reaches the desired height, then shut down the engine, change the pitch angle to horizontal and thrust up in order to get the desired speed. But I have no ideea how to do this.
clear; clc;
format longG
figure;
stage= [11600 4424 1400 275;
1650 695 300 37;
564.25 258.92 93.09 22.27;
47 37 36 28];
stage(3,:)=stage(3,:).*(10^3)
% Rocket parameters based on scout I row 1 gross mass; row 2 dry mass; row 3 engine thrust; row 4 burn time
% Isp=[280 22];
x=0;
z=0;
vx=0;
vz=0;
theta=90;t=0;deltat=0.25;
g=9.80665 ;
dth=-1.2;
K=(6.67428+0.00067)*10^(-11);
% Vertical Take off
while vz<=100 && z>=0
v=sqrt(vx^2+vz^2);
if(t<stage(4,1))
T=stage(3,1)
m=sum(stage(1,:))-(stage(1,1)-stage(2,1))/stage(4,1)*t
end
ax=1/m*T*cosd(theta);
G=m*g
az=1/m*(T*sind(theta)-G)
vx=vx+ax*deltat
vz=vz+az*deltat
x=x+vx*deltat;
z=z+vz*deltat;
theta=atan2d(vz,vx);
t=t+deltat;
plot(x,z,'r*'); hold on;
% pause(0.1)
end
% Establishing trajectory
;flight=0;
vz
for i=1:3
while t<stage(4,i);
v=sqrt(vx^2+vz^2);
T=stage(3,i)
m=sum(stage(1,i:end))-(stage(1,i)-stage(2,i))/stage(4,i)*t
% r=distance(x,z)
g=gravitacc(x,z)
G=m*g
ax=1/m*T*cosd(theta);
G=m*g
az=1/m*(T*sind(theta)-G);
vx=vx+ax*deltat
vz=vz+az*deltat
x=x+vx*deltat;
z=z+vz*deltat;
% theta=atan2d(vz,vx)+dth*deltat;
if theta>=15
theta=theta+dth*deltat;
end
t=t+deltat
plot(x,z,'r*'); hold on;
% pause(0.1)
end
plot(x,z,'bd');
text(x,z,num2str(i));
flight=flight+t
t=0;
end
t=0;
dth=-0.5
i=4
theta
while t<=600 & vz>0 & vx<=7800 & t<stage(4,i)
v=sqrt(vx^2+vz^2);
T=stage(3,i)/2.5;
m=sum(stage(1,i:end))-(stage(1,i)-stage(2,i))/stage(4,i)*t;
% r=distance(x,z)
g=gravitacc(x,z);
ax=1/m*T*cosd(theta);
G=m*g;
az=1/m*(T*sind(theta)-G);
vx=vx+ax*deltat;
vz=vz+az*deltat;
x=x+vx*deltat;
z=z+vz*deltat;
theta=atan2d(vz,vx);
if theta>=0 theta=theta+dth*deltat;
end
t=t+deltat;
plot(x,z,'r*'); hold on;
% text(x,z,num2str(vx));
% pause(0.1);
end
% theta
% m
while t<=600 & vz>0
v=sqrt(vx^2+vz^2);
T=0;
% m=sum(stage(1,i:end))-(stage(1,i)-stage(2,i))/stage(4,i)*t;
% r=distance(x,z)
g=gravitacc(x,z);
ax=1/m*T*cosd(theta);
G=m*g;
az=1/m*(T*sind(theta)-G);
vx=vx+ax*deltat;
vz=vz+az*deltat;
x=x+vx*deltat;
z=z+vz*deltat;
theta=atan2d(vz,vx);
t=t+deltat;
plot(x,z,'r*'); hold on;
% text(x,z,num2str(vx));
% pause(0.1);
end
% m
% theta
% fligh=flight+t
z
vx
vz
• KSP has no built in guidance at all. I'm guessing you're looking at MechJeb? – lamont Jan 13 at 3:43
• And the book "Orbital Mechanics for Engineering Students" has its code appendix up the publisher site: booksite.elsevier.com/9780080977478 and D.40 is "Calculation of a gravity-turn trajectory" which may be a better example for you to start with. – lamont Jan 13 at 3:45
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2019-06-19 23:52:55
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https://mathematik.univie.ac.at/en/research/seminars/seminars-at-the-kgrc/research-seminar-in-set-theory/full-news-display/news/generalised-descriptive-set-theory-part-ii/?no_cache=1
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# Generalised Descriptive Set Theory, part II
21.03.2023 15:00 - 16:30
M. Moreno (U Wien)
We have introduced the notions of $$\kappa$$-Borel class, $$\kappa$$-analytic class, $$\kappa$$-analytic-coanalytic class, $$\kappa$$-Borel* class in the previous talk. In descriptive set theory the Borel class, the analytic-coanalytic class, and the Borel* class are the same class, we showed that this doesn't hold in the generalized descriptive set theory.
In this talk, we will show the consistency of "$$\kappa$$-Borel* class is equal to the $$\kappa$$-analytic class". This was initially proved by Hyttinen and Weinstein (former Kulikov), under the assumption V=L. We will show a different proof that shows that this holds in L but also can be forced by a cofinality-preserving GCH-preserving forcing from a model of GCH, but also by a $$<\!\kappa$$-closed $$\kappa^+$$‑cc forcing.
Students at Uni Wien are required to attend in person.
Organiser:
Location:
#### SR 10, 1. Stock, Koling. 14-16, 1090 Wien
SR 10, 1. Stock, Koling. 14-16, 1090 Wien
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2023-03-21 04:21:26
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https://bioinformatics.stackexchange.com/questions/6906/how-to-score-how-densely-bases-are-distributed-along-a-given-oligo
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# How to score how densely bases are distributed along a given oligo?
Suppose I have a short chain string of oligos, and I want to rate them from 0 to 1 based on how clustered the "A"s are in respect to the chain. For example:
"AAAAACCCCC" #should score as 1 for being as clustered as it possibly could
"ACACACACAC" #should be scored as 0 for being as thinly distributed as it could
Current idea is to strip both ends of A's, then remove first and last characters, then
gaps = ''.join(str(x) for x in deck).split("A")
listofCs = [x for x in gaps if x]
to join it all together as a list of how big the gaps are (i.e. the "C"s), and simply count the average size
average = sum(map(len,listofCs))/len(listofCs)
But I feel like there should be a much more elegant solution to this. It feels messy. How can I improve the scoring?
• I would look at how much consecutive the different elements are in the sequence. But how would you rate a sequence like AACCAACACCAACC ? Is it for DNA or protein sequences (how much elements do you expect, 4, [5 with an N], 21) ?
– llrs
Jan 31, 2019 at 11:24
You could count the number of transitions between an A and a 'not A', and divide this number by the total number of As. You might also need to scale by the sequence length, if you have different length sequences that you want to compare.
Edit: also, there might be a correction to make if one or both ends of the sequence are an A
• Interestingly idea. Can you think of any edge cases that would cause problems for this scheme?
– Tom
Feb 2, 2019 at 2:35
• Zero As would throw a spanner in the works. Feb 4, 2019 at 13:51
The Markov chain approach is very interesting. You could deploy a maximum likelihood approach where instead of scoring the genetic distance within a homologue, the algorithm looks at its physically linked neighbour. The genetic distance of the transition point i.e. AAAA then CCCC is deducted from the final genetic distance so that AAAAACCCCCC would be zero. It would be fairly easy to script and use PAML or equivalent where the default phylogeny would be a star tree.
The challenge is to establish a Deep Learning approach. Its not impossible, you just assign pre-determined classes of oligos which fall into the definition you want and use these as your training set.
To define the base density score, or clustering score, of a sequence $$S$$ we can think of each position in the sequence as being part of a Markov process. For the sake of symmetry around a single base we consider odd-order Markov models from 3-onward where we ask what is the probability that we see the character at $$S_i$$ in the $$j$$ bases from $$[S_{i-j}, S_{i+j}]$$. For each column we have length(S) -1, $$|S|-1$$, possible probabilities to calculate per window.
We can't look left or right of the edges of the sequences so the probabilities are skewed, but we can get a rough idea of what is going on.
• For each base position we calculate a clustering score of $$\frac{1}{w}\Sigma_{j=1}^{j=w} -log_2 \frac{m(S_{i-j}, S_{i+j})-1 + \frac{1}{|S|}}{|S_{\{i-j, i+j\}}| -1 }$$ where:
• $$w$$ is the window size we can look at. The max is $$|S|-1$$ and the min is $$1$$. This just means that we can calculate a score in the context of the whole sequence or we maybe only care within some smaller window, like 5 bp flanking either side of the base in question.
• The function $$m(S_{i-j}, S_{i+j})$$ is just the number of counts of matching bases in the window $$j$$ on either side of the base $$S_i$$. For example in the sequence gatAgga, if $$S_i$$ is the uppercase A and $$j$$ is 3, the function $$m(S_{i-3}, S_{i+3})$$ returns 3, since there is the A base at $$S_i$$ and the two additional a bases at $$i-2$$ and at $$i+3$$. We have to subtract one from this number because we don't want to factor in the matching base at $$S_i$$.
• We then add $$\frac{1}{|S|}$$ to that number. This is a pseudocount that we need to add to prevent from taking the log of zero. This number is pretty arbitrary but I chose $$\frac{1}{|S|}$$ so that non-clusting regions' scores aren't too high.
• We then divide that numerator by the length of $$S$$ from the little window that we just looked at minus one, $$|S_{\{i-j, i+j\}}| -1$$. For example in the sequence gatAgga from above we only want to consider the bases gat gga. This is 6 bases.
• We then take the negative log to convert the proportional scale to a log scale.
• The summation just says that we do this for every symmetrical window size around the base in question then sum them up. For example we would calculate the ratio for tAg, atAgg, and gatAgga.
• The $$\frac{1}{w}$$ just gives the average log clustering score of all the windows that we looked at. One nice thing about this is that the impact of each increasing window size has less effect on the ratio, so there is a sort of convergence that happens.
Implemented in python:
from math import log as mlog
s = "GGATAGGACATTTTTTTTTTGAGCTTTTATTTTCCGA"
assert len(s) > 0
pseudocount = 1/len(s)
#maxdepth = 5
maxdepth = len(s)
score = []
for i in range(len(s)): # iterate positions
temp = []
for j in range(1, maxdepth): # iterate depth
rmin = max(i-j, 0)
rmax = i+j+1
num_matches_min_one = sum([1 for k in s[rmin:rmax] if k == s[i]])-1
len_s_min_one = len(s[rmin:rmax]) -1
sim = abs(-1 * mlog((num_matches_min_one/len_s_min_one) + pseudocount, 2))
#print("i={}, j={}, rmin={}, rmax={}, sim={}".format(i, j, rmin, rmax, sim))
temp.append(sim)
print("{0:d}\t{1}\t".format(i, s[i]), end='\t')
score.append(sum(temp)/len(temp))
print("{0:.4f}".format(score[-1]))
import matplotlib.pyplot as plt
fig, ax = plt.subplots()
x = [i+1 for i in range(len(s))]
plt.bar(x, score)
plt.xticks(x, [k for k in s])
ax.set_yscale("log", nonposy='clip')
plt.xlabel('sequence')
plt.ylabel('-log2(cluster-ness)')
plt.title("cluster-ness score")
plt.savefig('output.png', dpi=150)
This calculates the matrix, prints out a tab-delimited file of the position, base, and log clusting score, then saves a plot of the results.
For the test sequence GGATAGGACATTTTTTTTTTGAGCTTTTATTTTCCGA and using the whole sequence as context, I got the following plot:
As you can see, the log clustering score is low around the things that are cluster-y and high around things that aren't cluster-y.
Note: OP requested that the score be scaled from 0 to 1. I instead used the log scale to make visualization easier.
Warning 1: If you plan on doing this for large sequences, it is really important to limit the maxdepth, or $$w$$, parameter. The time complexity for this algorithm is $$O(|S|*w)$$, so the number of calculations can blow up to $$O(|S|^2)$$.
Warning 2: It is also important to limit the $$w$$ parameter because the edge effects propagate through the entire sequence when $$w=|S|$$. Look at these examples built off of OP's initial suggestions of AAAAACCCCC and ACACACACAC. The clustering score never stabilizes when $$w=|S|$$, however it does stabilize away from the edges when $$w=5$$.
For 50 characters of ACACAC...:
For 50 characters of [25xA][25xC]:
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2022-08-08 01:18:06
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https://wafflescrazypeanut.wordpress.com/2013/09/29/mirror-in-the-atomic-level/
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# Mirror in the atomic level…
Last week, I was trying to take a tour through these mirrors and reflecting stuff. But, I failed miserably. I thought of including other wonderful things that mirror can do, then explore deeply into the atomic level. Why I didn’t do that? I’ve already told you that I have some complicated & funny memory losses. When I was writing last post last time, I suffered a blackout which is the reason why it took so long for me, to post the “post”…
## Into the atomic level…
You say, “How exactly does reflection & refraction occur in the atomic level? I know it requires quantum electrodynamics. But, can’t that be explained in understandable terms?” Of course it can be explained. Before that, let me phrase that electromagnetic waves don’t have magnetic fields. You may well shout at me, “Are you drunk?”. No, but that’s sorta truth. Magnetic fields are a consequence of electric fields along with a pinch of special relativity. (Derek & Henry have teamed up to explain how ferromagnets and electromagnets work… Check those out!!!) So, it’s these electric fields we should be concentrating for now. Now that you’re cool, we can dig deeper into the subject. Two things happen in the mirror. First, the photons refract through glass, then they reflect at the metal-glass interface.
The common explanation goes like this. A light photon interacts with an electron in the atom of the material, by which it gets absorbed and then, re-emitted. The net effect of all these absorption & re-emission is what we perceive as reflection. This is a bit misleading. Because, this isn’t what’s really going on…
## Traveling through glass
Classical electrodynamics says this. The incoming photon can be thought to have wave-like behavior. Having oscillating electric fields, it interacts with the atomic structure, and oscillates (formally, it’s called polarization) the atomic dipoles back & forth. What I mean by dipole here? The atoms can be thought of to consist of a net positive charge bunch (which is the nucleus, of course) and a negative charge bunch (the surrounding electron cloud) which can be displaced slightly in the presence of an external electric field. But, this field is oscillating. So, the dipoles are left into tremendous oscillations which in turn, cause them to emit electromagnetic radiation (an accelerating charge radiates electromagnetic waves) of exactly the same frequency, but with a slight phase shift (which has two possibilities)…
• If there’s a shift of $\pi/2$, it corresponds to a refractive index of something greater than 1, which is what’s happening in materials like glass, the actual refraction. This is also the reason for light traveling slowly in these media.
• If the shift is a $\pi$, the waves destructively interfere and cancel out each other.
A phase shift somewhere between $\pi/2$ and $\pi$ corresponds to both reflection and absorption by some amount, and that explains the visible light frequencies for most materials.
Now, the macroscopic superposition (the total) of all these radiated waves, along with the original wave, gives you a summed up wave pattern that has experienced a net phase shift, which causes it to lag behind the incoming wave (which also explains the reduced speed). This whole mix up, is the observed beam of light that’s refracted and specular-reflected. Question: “Why does the frequency stays the same during these photonic interactions?” Think of this. As there’s no such thing as conservation of photon numbers, the photons can be destroyed and created. But, there’s a constraint. Either the photons are absorbed by the material (destructive interference) as a whole, or just transmitted (refracted or reflected) as a whole.
For example, in a dielectric material like glass (where there are no free electrons), the oscillating electric fields in the wave simply shakes the atomic dipoles back & forth, which causes them to radiate their own electromagnetic waves of the same frequency. “Why?” I’ve told you. Photons are either absorbed as a whole or just left out untouched. There’s no partial absorption, like 40% is sucked up from this photon’s body while 60% is left free (leading to the decrement in 40% of the energy, which is frequency). Meh… That’s nonsense..!!! Okay, now the forward radiation (a fancy name for the thing that’s going along the direction of the incoming wave) from these dipoles have a phase difference of $\pi/2$ with the incoming wave, which we perceive as the refracting beam. The backward radiation (going in the opposite direction) is simply the reflection (4% for glass) on such materials (like water, glass, etc.). “What about the sideward radiation?” They follow all sorts of complicated paths and cancel out one another. So, we don’t really have to worry about them…
In case of metals (which have valence electrons), the oscillation of the free electrons in the dipoles make the situation more complicated that they cause the dipoles to emit a forward radiation that has a phase difference of $\pi$, which leads to destructive interference, while the backward radiation is perceived as the reflected beam.
## Quantum laws run the whole movie…
Is that all? No, it’s confusing. How? One question. How do these dipoles exactly know where to put the photon back? We’re observing specular reflection everyday, and it happens as per the laws of reflection, that the angle of incidence equals the angle of reflection. So, how does it know where to emit the photon, while it’s busy whizzing around here and there in the lattice? It’s not just one. All the gazillion of the terrible things do the same. How is it that all those are synchronized so well?
Answer: The dipoles are not at all synchronized. The photon emission is a random process and it can be emitted anywhere, towards some angle which doesn’t agree with the law of reflection, even towards the source itself. Woah…
This issue was once addressed by Sixty Symbols (they’ve really tried very hard…)
This is where quantum electrodynamics rushes in. That’s why I told you to watch QED lectures last time. Anyways, the theory goes like this. Firstly, it just doesn’t make any sense to speak out that a photon interacts with a specific atom, nor it follows a definite path. It’s a quantum particle. It can interact with all the atoms, or take all the possible paths it can, and you can’t say anything about it. In the meantime, what we can do, is compute the chance (i.e) how likely the photons choose a specific path, after making a hell lot of observations on the occurrence with photo-multipliers and stuff. As far as I’ve seen, this is really a good explanation, based on the QED lectures, addressing Feynman’s path integral. I’m not going into it right now (as I’ve already suggested you to watch his lecture). But, this is what it roughly states…
We don’t go around defining a particular direction or path for the photon. As we can’t measure which atoms exactly the photons interact (i.e) where exactly they get scattered, we just calculate the superposition of all the possible outcomes. Though the probabilities of a specific photon interacting with individual atoms are small, the whole thing summed up for all the atoms in different layers of the material (different contributions) comes out to be a vast number. So, when we do this mathematical treatment, it just turns out that the the summed-up contributions of all possible paths of the photons is the path that has the shortest time period, which is the one that obeys $\theta_i=\theta_r$
It should be noted that this is a collective phenomena. I mean, there’s no such thing as, “a photon interacts with an electron“. It can interact with a lot of electrons, all at once. And, it does. But, you can’t observe it. The interactions lead to corresponding phase differences, that either add up or cancel out as a whole (interference). The whole business going on here, can be burned down to a single technical phrase. A photon follows a geodesic path, which simply is a straight line in a 2D flat space (which isn’t really necessary here, but I can’t resist).
Inspired by this post written by Chad Orzel (a professor) at Physics Stack Exchange.
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2017-02-24 17:39:19
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http://www.lecturematerials.co.uk/?x=y:06;m:10
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### Data Publishing
Tuesday, 31 October, 2006
The GRADE project, which I am a collaborative partner on, is concerned with scoping geospatial repositories. The project has principally been tackling legal and technical issues regarding their establishment and, I think, has made some very good progress. Yet behind all this work you do actually need people to deposit data for inclusion in a repository. And this is where the rub is. At the moment we have data centres (buzz word 5-10 years ago) and we are now seeing the increased establishment of institutional repositories. Yet what/where is the impetus for actually depositing data?? I suspect that this is partly subject specific. My impression is that subjects such as physics have a greater tendency to data share. In geosciences its usually a case of keeping what you have collected and only ever publishing the results; not the data itself. To be fair this is beginning to change with the research councils in the UK requiring the deposition of data from funded work. But how much data (from research) actually results from research council funding? My impression is less than half (although if anyone has any figures that would be interesting).
So we have the situation where there is a “top down” establishment of respostories, but no one is actually interested in using them. We have researchers collecting data (for research), but it is research publications that drives the agenda (NOT the data). I know that I see absolutely no reason why I should share primary data and, indeed, I like to discuss with people potential uses before sharing. Then of course we have the vested interests of the institutions that employ researchers. They are directly or indirectly funding much of this research and there is increased interest in “monitoring potential assets” (although quite to what extent institutions have a claim to IPR is another matter).
So where does that actually leave things?? Well Mahendra Mahey (at the GRADE meeting this week) provided a summary of repository work in the UK and (briefly) summarised some points that Pete Burnhill (Director of EDINA) was making along these lines. And that is that data should be published. As a community, academics need to be encouraged about the positive aspects of data sharing and see this as an opportunity to publish. Indeed one could argue that data publication should be seen as a valid publication route. And in the same way that journal articles are peer reviewed, so data should also be. This is a route that we have been toying with at the Journal of Maps. Several articles have data published with them (e.g. Stokes et al. They have been checked for appropriateness but not explicity reviewed in the same manner the article was. I am currently reviewing how useful this “service” is, with the potential to asks reviewers to comment on submitted data, as well as having a separate data reviewer. This actually raises a whole host of other questions concerning data preservation (as opposed to a repository) which I won’t comment on at this moment.
With the above comments, I think it is clear that I’m in favour of data publication, but I am inclined to think at the moment that the data should follow the research (hence the reason for publishing the data with the article at the Journal of Maps). The problem with separating data and content is that maintaining the explicit link between the two becomes more complex (just look at journals from the 19th century to see how effective immediacy is). It also makes the peer review process much simpler. That isn’t to say that data can’t be stored in a repository, but that, in the first instance, it might be better placed with the article. Indeed, I could see the research councils requirement for copies of publications and data deposition taken a stage further and requiring research articles to have data published with them. Clearly the emphasis is then shifted to the journals many of whom will not be placed to deal with it. However the whole research publication ethos is changing (e.g. open access) and it is time that journals become proactive. Indeed, with Wiley and Elsevier being so prominent (and supporting things like permanent electronic archives), it would only require these two organisations to support such an initiative for it to really take off. Whilst in principle it sounds a reasonable idea, there are many barriers. Not least the sheer volume of some data sets within a web based infrastructure where most journals struggle to offer more than a static PDF.
### More on TOIDs
Monday, 30 October, 2006
Following on from my earlier TOID blog, I was at EDINA (University of Edinburgh) where I was briefly chatting about the Mastermap trials. Getting on to the subject of TOIDs I mentioned about TOID searching and, I am happy to report, EDINA will be incorporating TOID searches in to the Digimap interface (although not just yet). So at least someone is doing it which is good to see. Interestingly, they are also working on exploiting the royalty free status of TOIDs. If you interested in some processing that has been performed, you could essentially be given a macro that list the TOIDs and actions performed on them. It is this file that could be stored (in a repository for instance) and emailed out and then you could legitimally obtain a licensed copy of the data and perform exactly the same operations. No need for the (intermediate) processed data set. Nice idea and I’ll be interested to see it in action.
### Palm Opera
Saturday, 28 October, 2006
Whilst I’m on the theme of playing with my Palm, I saw the recent release of Opera Mini for the Palm (and this interested me because WebPro that shipped with my Palm is not great). Opera for a long time was “the” alternative to Internet Explorer. Whilst it was a paid-for app, it was for small and fast, particularly suited for optimising browsing over dial-up connections. It also was the first browser (I think) to introduce an MDI (multiple document interface) based around tabs. Anyway, the release of Firefox eclipsed many of the in roads Opera had been making. To start with Opera went advertising supported and then ultimately free. However where Opera has been making big in-roads is in the mobile market. Its ability to design small and powerful browsers is ideally suited to this area and it has versions of Opera Mini running on a variety of operating systems. PalmOS has, until recently, not been one of those supported, but the moderate success (particularly stateside) of the Treo smartphones has led to a release which was recently upgraded to version 2.
Opera Mini is actually a highly optimised (read: fast) Java applet that runs on top of IBM Websphere. You need to install this before Opera Mini. Fire it up and everything just works. Opera has done a good job of using screen space well and it succesfully (via a proxy) loaded pretty much any website I threw at it (ebay, GMail, Google etc etc). But the screen is small so don’t expect to got lots of productivity out of it. Like anything, its great to have Office, VNC, email and web all on the Palm, but only when you are in a tight spot.
Tuesday, 24 October, 2006
Google Maps was recently released for the Palm platform so I wandered on over to download it. Whilst Palm OS as a platform for PDAs is dwindling (but has a fantastic software base), Palm has been moderately successful with its range of Treo smartphones. Hence a Palm OS version of Google Maps. And I was astounded at how truly powerful an application it is. The interface is simple and offers three main functions:
• 1. Maps
• 2. Satellite images
• 3. Route planning
And what it does it does very well. Zooming in is quick (although whilst you can pan you can’t interactively zoom) and refresh rates fast. You can switch between map and satellite view. I found this strangely addictive and found the whole experience of interactively using 15cm digital aerial imagery “on demand” truly amazing. Finally the route planning is simplicity itself and very effective at showing the route.
### LaTeX: Abiword
Monday, 23 October, 2006
LaTeX is a mark-up “language” that users learn in order to write LaTeX documents. If you have ever done any HTML by hand then it is similar. The markup accesses the background macros that control all the layout. Below is an example of a LaTeX document:
\documentclass[12pt,a4paper]{article}\usepackage[pdftex]{graphicx}\usepackage{multicol}\begin{document}\section*{Notes on my new paper}Some notes simply typed in to the documen. I can also add some \textbf{bold} and \textit{italic}.\end{document}
So nothing desparately exciting or difficult, althought laying out graphics and tables can get quite fiddly. But of course remember that this is a typesetting program. Not a DTP one.
If this is all a bit of a large learning curve in the first instance then AbiWord is a good place to start. It is a cross-platform WYSIWYG word processor that is pretty good. It doesn’t have all the bells and whistles of Microsoft Word, but does do nearly everything you want and pretty well. Its also OpenSource (and if you want a portable version, pop over to the people at PortableApps). What makes AbiWord stand out a little more is (via plugins) its support for MS Word import (and others) and LaTeX export. It also has a pretty nifty equation editor that utilises an implementation of the LaTeX equation language. So you can import Word documents, save them as LaTeX files and run them through pdftex to create LaTeX PDFs. All very neat.
### LaTeX Musings
Friday, 20 October, 2006
Before I get in to this blog, I should note that this will form one of several entries on LaTeX.
When we first started the Journal of Maps we decided to typeset the material ourselves and pondered for quite a while about which software to use. Whilst DTP software first sprang to ming (e.g. InDesign), these are not wholly designed for free flowing text, but rather short, styled, pieces. What we really wanted was software for typesetting and some hunting around the internet pointed me in the direction of LaTeX. This is a version of TeX which adds many macros to allow “ordinary” users to access the true typesetting power of TeX. Tex is nicely summarised by Wikipedia as “a typesetting system created by Donald Knuth. Together with the METAFONT language for font description and the Computer Modern typeface, it was designed with two main goals in mind: first, to allow anybody to produce high-quality books using a reasonable amount of effort, and, second, to provide a system that would give the exact same results on all computers, now and in the future.”
The source code is in the public domain and there are serveral versions available across many platforms. Whilst for typesetting the output is excellent, because TeX will remain essentially unaltered, code from 1985 should work without problems in 2085. One only has to look at early DOS Word Processor programs, or indeed the BBCs Doomsday Project, to see how quickly file formats can date. This shouldn’t be an issue with TeX.
So at JoM we had settled on the software to use. We then had to pick a distribution, platform and design a template. To keep things short for the moment, I’ll just note that we use Windows XP and settled on LaTeX, which meant going for a version called MiKTeX. And this does everything we want it to!
### Latte Heaven
Tuesday, 17 October, 2006
Bit of a plug for one of my local coffee houses, Latte Heaven. Great coffee (using a blend of their own, roasted in Leighton Buzzard), a good menu of food and a nice place to chill out. The website is, well, not very good. It is template driven though and, amusingly, gives a link to the control panel for the site at the web hosts. Funny what people leave online.
Anyway, if you’re passing through Dunstable, give it crack. You won’t be disappointed.
### BBC News: The map gap
Monday, 16 October, 2006
BBC News had an interesting article today entitled The map gap. It covers familiar catographic ground on how we actually represent an ellipsoidal Earth on a flat piece of paper (Nice quote from Steve Chilton: “If you peel an orange, you can’t lay it flat and there’s never an answer to that”); first year geographers, this is exactly why we have a lecture on map projections and coordinate systems! It then briefly introduces Google Earth (and ilk), mapping websites and the whole idea of “user-data” (e.g. OpenStreetMap) and mash-ups. Whilst the article itself doesn’t break any new ground, it is interesting in that it places cartography at the centre of these developments and subtlely (or unintentionally?) asks how the subject can respond to such rapid changes. And this is a good question; will cartography reduce to a niche subject as spatial data users continue to make visually poor (or wrong!) maps or can it re-invent itself? It also demonstrates how mainstream spatial data, visualisation and re-use (mash-ups) have become.
### Searching TOIDs?
Wednesday, 11 October, 2006
Following on from my initial thoughts on Mastermap, I was having a ponder about the use of TOIDs. TOIDs essentially represent another addressing system for the UK, allowing identification at the object level. And of course, following on from my blog on AGI Tat, the OS are advertising TOIDs as royalty free. So whilst they can be used for much more than simply addressing, are there any searchable interfaces based on TOIDs (i.e. web based)? Anyone know??
P.S. I can be TOIDed at 1000041424855 (but does anyone know where that actually is!)
### Mastermap Styling
Thursday, 5 October, 2006
With EDINA preparing OS Mastermap data for general distribution as part of Digimap, one of my colleagues has become an early adopter for trialling purposes. Whilst I have come across Mastermap in various guises, this was the first time I had seen the whole processing side up close. My colleague downloaded a couple of layers for a 5×5km area which came to over 200Mb. The first problem was loading it in to ArcMap which is one of the least standards compliant GIS packages around (although this might well be coming in v9.2). A trip over to ESRI UK and a download of MapManager 9 allowed the conversion of the OS GML into a geodatabase. Mastermap then loaded fine, although with default ESRI symbolisation. A lot of digging and we finally came across a style file that symbolises the Mastermap data in the same fashion as the default OS styling (and pleasant enough it is too). Its worth noting that you can get a free OS Mastermap GML Viewer from Snowflake Software.
We then downloaded all layers (although no imagery) for a 5×5km area and, 2-hours later, MapManager produced a 750Mb geodatabase. Hmmm, some planning ahead me-thinks. This is clearly a big headache for EDINA (it took about 6-hours for the data request to be processed, rather than the 2-3 minutes for LandLine), hence the need for testing!
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2019-12-15 08:30:32
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https://mathoverflow.net/questions/238942/chief-factors-and-local-formation
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# Chief factors and local formation
Every thing below is concerned with finite groups.
A class of groups is a collection $\mathcal{X}$ of groups with the property that if $G \in \mathcal{X}$ and if $H \cong G$, then $H \in \mathcal{X}$.
A class of groups $\mathcal{F}$ is called a formation, provided that the following conditions are satisfied:
(1) If $G \in \mathcal{F}$, then $G/N \in \mathcal{F}$ for any normal subgroup $N$ of $G$,
(2) If $G/M$ and $G/N$ are both in $\mathcal{F}$, then $G/(M \cap N) \in \mathcal{F}$ for any normal subgroups $M$ and $N$ of $G$.
Now let $\Pi$ be the set of all prime numbers. Then, by a formation function $f$, we mean a function $f$ define on $\Pi$ such that $f(P)$, possibly empty, is a formation. A chief factor $H/K$ of a group $G$ is called $f$-central in $G$ if $G/C_{G}(H/K) \in f(p)$ for all primes $p$ dividing $|H/K|$. A formation is a local formation if there exists a formation function $f$ such that $\mathcal{F}$ is a local formation defined by $f$ and we write $\mathcal{F}=LF(f)$ and call $f$ a local definition of $\mathcal{F}$.
Among all the possible local definitions for a local formation , there exists exactly one of them, denoted it by $F$, such that $F$ is integrated (i.e., $F(p) \subseteq \mathcal{F}$ for all $p \in \Pi$) and full (i.e., $\mathcal{S}_{p}F(p) = F(p)$ for all $p \in \Pi$) where $\mathcal{S}_{p}$ is the class of $p$-groups.
How did the author used Lemma 2.3 (see the lemma)
in this portion of the proof of Corollary 3.2. (see the portion)?
My question is exactly in the line that is underlined with blue. How did the authors conclude that $G/C_{G}(K_{1}/K_{2}) \in F_{1}(q)$?
I know that one possible local definition for the class of $p$-nilpotent groups, denoted by $\mathcal {N}_{p}$, is the following: We define the formation function as follows: $f(p)=(1)$ and $f(q)=\mathcal {G}$ for $q \neq p$, Where $(1)$ is the class that contains only the trivial group $\{1\}$ and $\mathcal{G}$ is the class of all finite groups.
According to the above local definition of $p$-nilpotent groups $G/C_{G}(K_{1}/K_{2}) \in f(q)$ clearly as $f(q)= \mathcal{G}$ the class of all finite groups. But the problem is that is this local definition is not integrated as $\mathcal{G}$ is not contained in the class of $p$-nilpotent groups $\mathcal {N}_{p}$.
I think that the full and integrated formation function (which is unique) $F_{1}$ for the class of $p$-nilpotent $\mathcal {N}_{p}$ groups is given by $F_{1}(p)=\mathcal{S}_{p}$ and $F_{1}(q)=\mathcal {N}_{p}$ for $q \neq p$, Where ${S}_{p}$ is the class of all finite $p$-groups and $\mathcal {N}_{p}$ is the class of all $p$-nilpotent groups. However I do not know how to prove that $G/C_{G}(K_{1}/K_{2})$ is $p$-nilpotent so that $G/C_{G}(K_{1}/K_{2}) \in \mathcal {N}_{p} = F_{1}(q)$.
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2019-08-24 10:31:45
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https://www.hackmath.net/en/math-problem/14851
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Weigh in total
I put 3/5 kg of grapes into a box which is 1/4kg in weight. How many kilograms do the grapes and the box weigh in total?
Result
m = 0.85 kg
Solution:
$m_{1}=\dfrac{ 3 }{ 5 }=0.6 \ \text{kg} \ \\ m_{2}=\dfrac{ 1 }{ 4 }=0.25 \ \text{kg} \ \\ \ \\ m=m_{1}+m_{2}=0.6+0.25=\dfrac{ 17 }{ 20 }=0.85 \ \text{kg}$
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2020-04-03 17:58:18
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## anonymous one year ago find all value of m such that sqrt(x) = m(x-2) has a solution greater than 4
• This Question is Closed
1. dan815
m>4 right
2. Empty
Square both sides of the equation I guess
3. dan815
are u looking for int solutions?
4. anonymous
@dan815 no, real numbers solution
5. anonymous
@Empty I guess you could do so. I've attempted and with a help with wolfram alpha, it got 0 < m < 1. But the arithmetic is just way tedious.
6. dan815
ok well look at the behavior of x/(x-2)^2, find its critical, and inflection points also intersect it with y=4^2 LINE
7. anonymous
just wondering if there is any other easier method
8. Empty
Solve for x, then set that value to be greater than 4. Hmmm Ohhh so you're looking for a trick, like a nice way to do it ok that makes this much more fun hahah ok hmmm give me a minute.
9. dan815
oh so x>4 not m
10. Empty
Just plug in x=4 and solve for m=1. Then you know m has to be greater than that, cause the square root is always positive, done.
11. anonymous
@dan815 yeah, it's about m that makes x > 4 as solution
12. dan815
you can also think about it like a sqrt function intersecting with a line displaced 2 units right, and m is the slope
13. anonymous
@Empty huhm... so sqrt(4) = m(4-2), which gives m = 1. And then?
14. dan815
|dw:1436389702432:dw|
15. Empty
You want all solutions larger than that, so m>1. That's all your solutions. Here look: https://www.desmos.com/calculator/ryk0xxgqmf
16. Empty
The case you solved for is the lowest possible case for the value of m. Well technically it's not because it's really m>1 not $$m \ge 1$$
17. dan815
ya thatsa the simplest way
18. anonymous
@dan815 I thought of it that way too.
19. dan815
just set x=4 and see what m has to be
20. dan815
m has to be less than that value
21. anonymous
@Empty well, set = m = 1.1, then x is about 3.764 which is not greater 4
22. dan815
he meant less
23. dan815
m >0 and less than 1
24. anonymous
Yeah, as I said earlier, 0 < m < 1 but this is with the help of wolfram alpha.
25. dan815
as a slope below 0 will make it intersect behind
26. Empty
Right right sorry I am backwards in what I said
27. Empty
XD haha sorry to be confusing
28. Empty
I was imagining as m gets smaller the value of the point of their intersection increases.
29. anonymous
yeah, that helps confirm our intuition but as far as the algebra go, is there a nicer way to get 0 < m < 1?
30. anonymous
The tedious algebra is solving this, which wolfram alpha did https://www.wolframalpha.com/input/?i=%28%284m%5E2%2B1%29+%2B+sqrt%5B+%284m%5E2%2B1%29%5E2+-+4%28m%5E2%29%284m%5E2%29%5D+%29+%2F+%282m%5E2%29+%3E+4
31. anonymous
I'm about to go soon. So just work on it if you're interested.
32. anonymous
find all value of m such that sqrt(x) = m(x-2) has a solution greater than 4 $$\sqrt{x}=m(x-2)\\x=m^2(x-2)^2\\x=m^2(x^2-2x-4)\\m^2x^2-(2m^2+1)x-4=0\\x=\frac{2m^2+1\pm\sqrt{(2m^2+1)^2+16m^2}}{2m^2}$$to prove *a* solution greater than $$4$$ all we need is:\frac{2m^2+1+\sqrt{(2m^2+1)^2+16m^2}}{2m^2}\ge4\\2m^2+1+\sqrt{(2m^2+1)^2+16m^2}\ge8m^2\\\sqrt{(2m^2+1)^2+16m^2}\ge6m^2-1\$$2m^2+1)^2 +16m^2\ge(6m^2-1)^2\\4m^4+4m^2+1+16m^2\ge 36m^4-12m^2+1\\32m^2\ge32m^4\\m^2(m^2-1)\le 0so our only solutions are where \(m^2=0$$ or $$m^2\le 1$$, which corresponds to $$|m|\le 1$$
33. anonymous
now, of course, some of these solutions are extraneous, since $$\sqrt{x}=m(x-2)\ge 0$$ so $$m<0$$ requires that $$x-2<0\implies x<2$$ however we want solutions $$x\ge4$$ we actually must have $$m>0$$ so $$0\le m\le 1$$
34. anonymous
now, if you wanted to guarantee $$x>4$$ strictly rather than $$x\ge 4$$ then the solution is really $$[0,1)$$
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2017-01-17 19:40:18
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https://gowers.wordpress.com/2008/02/01/removing-the-magic-from-stirlings-formula/
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## Removing the magic from Stirling’s formula
Another topic on the syllabus for the probability course I am giving is Stirling’s formula. This was lectured to me when I was an undergraduate but I had long since forgotten the proof completely. In fact, I’d even forgotten the precise statement, so I had some mugging up to do. It turned out to be hard to find a proof that didn’t look magic: the arguments had a “consider the following formula, do such-and-such, observe that this is less than that, bingo it drops out” sort of flavour. I find magic proofs impossible to remember, so I was forced to think a little. As a result, I came up with an argument that was mostly fully motivated, but there is one part that I still find mysterious. On the other hand, it looks so natural that I’m sure somebody can help me find a good explanation for why it works. When I say “I came up with an argument” what I mean is that I came up with a way of presenting an existing argument that doesn’t involve the pulling of rabbits out of hats, except in the place I’m about to discuss.
The first idea in proving Stirling’s formula is a very natural one: estimate $\log(n!)$ instead. Since that equals $\log 2+\dots+\log n$, it is natural to approximate the sum by the integral $\int_1^n\log x dx$. However, if you do that then the error will be around $\log n$. This gives you a good rough-and-ready approximation to $n!$ but nothing like as good as Stirling’s formula.
The next idea is to turn things round and try to approximate the integral by a sum rather than the other way round. The obvious next approximation to try is approximating $\log x$ by the function $g(x)$, where this is defined to equal $\log x$ when $x$ is an integer and to be linear in between consecutive integers. Then you can expect a much smaller error, and indeed it is not hard to prove that the difference between the integrals of the two functions is bounded. (The proof I gave for this dropped out rather nicely: if $r$ is a positive integer then by looking at derivatives it is easy to show that between $r$ and $r+1$ we have the inequalities $\log r+(x-r)/(r+1)\leq g(x)\leq\log x\leq\log r+(x-r)/r$. Therefore, the difference between the integrals of $\log x$ and $g(x)$ between $r$ and $r+1$ is at most $(1/r-1/(r+1))/2$. This gives a nice telescoping sum as an upper bound for the sum of all those areas.)
Another important observation is that the integral of $g(x)$ between 1 and $n$ works out to be $\log(n!)-\log n/2$ (because the integral between $r$ and $r+1$ is $(\log r+\log(r+1))/2$). Thus, we have a rather close relationship between $\log(n!)$ and an integral that we can calculate: $\int_1^n\log x dx=n\log n-n+1$.
It is not hard to see that this argument shows that the ratio of $n!$ to $(n/e)^n\sqrt n$ tends to a limit. The one remaining magic part is the calculation of that limit. To do this, we define $c_n$ to be the integral $\int_1^n(\log x-g(x))dx$. We have established that the sequence $(c_n)$ converges; our problem is equivalent to working out what the limit is. Here there is a very nice trick, which is to consider instead the sequence $a_n=2c_n-c_{2n}$, which clearly converges to the same limit. But if you calculate $a_n$ you find that you can get a very good estimate for it if and only if you can get a very good estimate for $\binom{2n}{n}$, essentially because inside what you write for $a_n$ you get $\log(2n!)-2\log(n!)$, which is $\log \binom{2n}{n}$. This trick is I think reasonably non-magic: if you know in advance that you can calculate $\binom{2n}{n}$ very accurately, then it is natural to try to use that information, and there is essentially only one way of doing that.
Even the fact that one can get a very good handle on $\binom{2n}{n}$ is not as magic as all that, since it is equivalent to estimating the probability that a subset of a set of size $2n$ has exactly $n$ elements. And that one can think about in all sorts of ways. For example, one can use the central limit theorem to estimate the probability that a random set has size between $n-c\sqrt n$ and $n+c\sqrt n$ and an elementary argument to show that the probabilities of the sizes within that range are all roughly equal.
However, we have not yet got to the central limit theorem in the course, so at this point I copied another proof, which goes roughly like this. You look at the integrals $I_n=\int_0^{\pi/2}\cos^n(x)dx$. By integrating by parts in a standard way, you can get the recurrence $nI_n=(n-1)I_{n-2}$. Since the $I_n$ clearly decrease, we get from this that the ratio of $I_{2n}$ to $I_{2n-1}$ tends to 1. But we can also give an exact formula for this ratio by using the recurrence and calculations of $I_0$ and $I_1$. If you do this and rearrange things a bit, then you find that you can show that $2^{-2n}\sqrt n\binom{2n}{n}$ tends to $\pi^{-1/2}$.
Finally, my question: is it just a coincidence that looking at the ratio of $I_{2n}$ to $I_{2n-1}$ gives you an expression that involves $\binom{2n}{n}$? Surely it isn’t, but for the moment this part of the argument still looks to me like a convenient piece of magic. It was clean enough to be easily memorable, so this did not present a problem for the lecture, but I’d still like to know why those integrals are so closely related to the binomial coefficient.
It would be remiss of me not to include a statement of Stirling’s formula in this account of the proof: it says that the ratio of $n!$ to $\sqrt{2\pi n}(n/e)^n$ tends to 1.
### 28 Responses to “Removing the magic from Stirling’s formula”
1. Pseudo-Polymath » Blog Archive » Friday Highlights Says:
[…] Maths, a proof. […]
2. Lior Says:
Hello,
It is standard to get Stirling’s formula by interpreting the sum $\sum_k \log k$ as a quadrature rule. You preferred the trapezium rule, but a slightly simpler variant is to interpret it as the midpoint rule for the interal $\int_{1/2}^{n+1/2} \log x\, dx$. In both cases, the error on each interval ($\null [k-1/2,k+1/2]$ or $\null [k,k+1]$ is bounded by a constant time the maximum of the second derivative of the integrand (times the cube of the interval length, but that’s a constant here), which here is summable over $k$. This is “why” this works, indepedently of the “magic” of the telescoping series you get from the trapezium rule.
The statement $4^{-n}\sqrt{n}\binom{2n}{n} \to \pi^{-1/2}$ is a variant of Wallis’s formula, for which you give the usual proof. Note that you are missing the exponent $n$ in the definition of $I_n$.
• M Says:
Here is a simple proof of Wallis and Stirling formulas. Email me for more pictures and details.
Consider a {\it one-dimensional random walk} of a point $\cal Q$
that starts at $x=0$ and at each step jumps from the point it
occupies to either right or
The probability of finding point $\cal P$ in
location $x=K$ after $N$ steps from its initial position is
given by the binomial distribution
$${P}(K,N)=\begin{cases} \dfrac{N !} {2^N\; \left(\frac{N+ K }{2}!\right) \left(\frac{N- K }{2}!\right) }, \text{ if }\vert K\vert \le N; \; N-K \text{ is even},\\ 0, \text{ otherwise}.\end{cases} \eqno{\rm (3)}$$
Let us now assume that
$$N=2n \text{ is even } \eqno{\rm (4a)}$$
and $\varepsilon$ satisfies
$$0<\varepsilon<\dfrac 5{30}. \eqno{\rm (4b)}$$
Formula (3) can be rewritten as
$$\hskip-0.2cm {P}(-2k,2n)= {P}(2k,2n)=\begin{cases} \dfrac{(2n) !} {2^{2n}\; \left(n!\right)^2 }\cdot \dfrac{\left(n!\right)^2 } { (n+ k)! (n- k)! }=\\ \hskip0.5cm \dfrac{(2n) !} {2^{2n}\; \left(n!\right)^2 } \prod\limits_{j=1}^{k} \dfrac{\;\;n-\ (k-j) \;\;}{n+ { j} }, \text{ if } 0\le k\le n,\\ 0, \text{ otherwise}.\end{cases} {\rm (5)}$$
Formula (5) implies that sufficiently large $n$
$$1-\sum\limits_{\vert k\vert < n^{0.5+\varepsilon} } P\big(2k, 2n \big) =\sum\limits_{\vert k\vert \ge n^{0.5+\varepsilon} } P\big(2k, 2n \big)\le 2^{- n^{2\varepsilon}} \eqno{\rm (6)}$$
and for $\vert k\vert \le n^{0.5+\varepsilon}$
$$\frac 1{\sqrt{ n}} e^{-\frac{k^2}{n} -8 n^{-0.5+3\varepsilon} } \le\frac {2^{2n}\; \left(n!\right)^2 }{(2n) !\sqrt{ n}} \;\; P(2k,2n) \le \frac 1{\sqrt{ n}} e^{-\frac{k^2}{n} +8 n^{-0.5+3\varepsilon} }.\eqno{\rm (7)}$$
The proof of formulas (6,7) is given in the Appendix.
Formula (7) implies
$$\bigg[ \frac 1{\sqrt{ n}}\sum\limits_{ \vert k \vert \le n^{0.5+\varepsilon} } e^{-\frac{k^2}{n}}\bigg] e^{ -8 n^{-0.5+3\varepsilon} } \le\frac {2^{2n}\; \left(n!\right)^2 }{(2n) !\sqrt{ n}} { \sum\limits_{\vert k \vert \le n^{0.5+\varepsilon}}P(2k,2n)} \le \bigg[ \frac 1{\sqrt{ n}}\sum\limits_{ \vert k \vert \le n^{0.5+\varepsilon} } e^{-\frac{k^2}{n}}\bigg] e^{ 8 n^{-0.5+3\varepsilon} }.$$
\color{black} Taking now limit as $n\to +\infty$ and using
$\lim\limits_{n\to +\infty}e^{ 8 n^{-0.5+3\varepsilon} }=\lim\limits_{n\to +\infty}e^{- 8n^{-0.5+3\varepsilon} }=1$,
$\lim\limits_{n\to +\infty} \sum\limits_{\vert k \vert \le n^{0.5+\varepsilon}}P(2k,2n)=1$ due to (6), $\lim\limits_{N\to +\infty}\dfrac 1{\sqrt{n}}\sum\limits_{{\vert k \vert \le n^{0.5+\varepsilon},} \atop{k \text{ is even}}} e^{-\frac{k^2}{n}}=\int\limits_{-\infty}^{+\infty}e^{-t^2}dt=\sqrt{\pi}$
we obtain (1).
3. Lior Says:
Oops … it seems I miffed the WordPress LaTeX syntax. I wish there was a “preview” button for comments …
4. gowers Says:
Thanks for that, and I’ve now sorted out your LaTeX and my wrong definition of $I_n$. I should have added that the magic of the telescoping sum wasn’t critical to getting the estimate to work, but it’s quite pleasant and the proof works for an arbitrary function with decreasing derivative. So that piece of “magic” doesn’t concern me.
5. David Speyer Says:
I think that, if you follow the approach by taking logs, you will always have to do something tricky to get the value of the constant. After all, for many sums, such as sum 1/n, sum 1/(n log n) or sum n log n, it is easy to get an asymtopic formula where the constant term is given by a convergent sum, but it is very rare that this constant has a closed form.
Here is a different approach where the constant comes out relatively naturally. The first step is less motivated than in your argument but, after that, everything feels natural to me.
$n!=\int_0^{\infty} e^{-x} x^n dx$
Now, a quick computation with derivatives shows that $e^{-x} x^n$ is optimized at x=n, so lets make a change of variables to put x=n at the origin. Set $x=u+n$, so
$n!=n^n e^{-n} \int_{-n}^{\infty} e^{-u} (1+u/n)^n dx$
The formula $(1+u/n)^n$ is screaming to be replaced by $e^u$. Just making this replacement gives us far too much error, so keep one more term.
$(1+u/n)^n=\exp(n \log(1+u/n))=\exp(u-u^2/2n+O(u^3/n^2))$
so the integrand is
$\int e^{-u^2/n+O(u^3/n^2)} du.$
Intuitively, we are done at this point, because $\int e^{-u^2/n} du=\sqrt{2 \pi n}$. Being careful, we have the problem that the error term $O(u^3/n^2)$ will overwhelm the terms we care about when u is too large. So the careful way to do this is to chop our integrand at $|u|=n^{a}$ for some $a$ between 1/2 and 2/3, and use some straightforward bounds to show that those tails don’t contribute.
6. David Speyer Says:
Stupid typos in the above:
The formula that doesn’t parse is
$n!=n^n e^{-n} \int_{-n}^{\infty} e^{-u} (1+u/n)^n du$.
I dropped the coefficient 1/2 in the Taylor series
$n \log(1+u/n)=u-u^2/2 n+O(u^3/n^2)$, and in several of the following formulas.
[Many thanks for the comment, and I’ve fixed the bugs now.]
7. Old Fart Says:
What’s wrong with magic? I was shown the summed up logs thing in a chemistry 1B lecture, probably before you were born, that was enough to crack out the boltzman distribution. My ‘please sir, what about the pis’ was rather dismissed as taking the piss. Beyond that, Copson’s ‘Theory of Functions of a Complex Variable’, Section 9.53 does the business in an old school way. Dave MacKay’s take (Information Theory, Inference and Learning Algorithms, page 2!) has a shade of the darker arts (including the central limit theorem) to it. None of these is maths, perhaps; Stirling’s theorem does, nonetheless, belong to the world.
8. David Watkins Says:
I found that proof a wonderful read. I stumbled across this blog not expecting to understand anything, so I am prompted to to write this. I like your style of “removing the magic”, seeing how it really works.
Thanks.
9. toomuchcoffeeman Says:
Shouldn’t the integral defining $I_n$ be taken from $\null 0$ to $\pi/2$?
Also (though in some sense this ties in with Lior’s post above) one non-magic reason to get ${2n\choose n}$ out of these integrals is that it is the constant coefficient in $(x+x^{-1})^{2n}$ — a fairly natural trick to try — so putting $x=e^{i\theta}$ and rescaling one gets
$2^{-2n}{2n\choose n} \frac{\pi/2} = \int_0^{\pi/2} \cos^n \theta\,d\theta$
I can’t see how to get round using (the proof of) Wallis’ formula, though.
10. toomuchcoffeeman Says:
Oops, bodged LaTeX. “No formula provided” should be 0, of course … and the last formula should be
$2^{-2n}{2n\choose n}\frac{\pi}{2} = \int_0^{\pi/2}\cos^n\theta\,d\theta$
[Many thanks for the comment, and I like the explanation for the appearance of the binomial
coefficient even if it doesn’t get the whole way to answering the question. I’ve corrected the bounds
in the integrals and fixed your LaTeX bug now.
]
11. B Says:
This is certainly pulling a rabbit out of a hat, but what about integrating the logarithm of the sine product of sin(pi*x)/pi*x from 0 to 1? This is, if nothing else, a fairly natural thing to do once you know about the sine product, and the proof comes out without much trouble. Certainly it isn’t hard to remember as a proof, at any rate.
12. gowers Says:
Just realized that the beautiful argument given by David Speyer a few comments up is almost certainly essentially the same as the argument I alluded to using the central limit theorem: your/his main steps are to centre and rescale a distribution and use the Taylor expansion up to the second term to get a Gaussian coming out, just as one does in proving the central limit theorem itself. But the fact that it comes out quickly and directly like that is very nice. And I myself find the first step pretty natural — if you’ve got the integral representation of the gamma function, then why not try to use it?
13. David Speyer Says:
I just realized this morning that the resemblance of my argument to the central limit formula is more than superficial: Let $F(x)=e^{-x}$ for $x \geq 0$ and $F(x)=0$ for $x < 0$. Then the $(n+1)$-fold convolution of $F$ with itself is $x^n e^{-x}/n!$ for $x \geq 0$ (and, of course, zero for negative $x$). So what I’m doing is basically verifying the central limit theorem by hand for independent eponentially distibuted random variables.
14. Nilay Says:
I tried out the integration $\int_1^n\log x dx$. This evaluates to $n log n - n +1$. This would mean that this approximation is off by $n^{\frac{1}{2}}$.
15. James Says:
Hi,
I was wondering if you’d mind if I used your blog article on Stirling’s formula for the next edition of “Eureka” (the journal of the Archimedeans undergrad. maths society here at Cambridge). It would most likely reach quite a broad audience of first years next year if you would consent. It might be worth appending David Speyer’s remarks also.
Of course if there is something else you’d prefer I’m open to suggestions.
Contact me at archim-eureka-editor@srcf.ucam.org if you are interested.
16. TSM Says:
Speaking of rabbits, here is a reference:
Strange Curves, Counting Rabbits, and Other Mathematical Explorations,
Keith Ball, Princeton Univ Press, 2003.
Chapter 6 (pp 109-126) is an excellent, motivated, wonderfully illustrated expository account of the path traveled largely by Gowers to Stirling’s formula. The Central Limit Theorem is discussed in the previous chapter, and so the estimate of the middle binomial coefficient 2n choose n is available. His approach to Central Limit Theorem is also excellent, starting from coin-tossing, and leading to the standard improper double integral exploited for the normal distribution. Explicit use of the big-Oh notation is avoided.
Ball also mentions that historically this approach is backwards; the Central Limit Theorem was first established using Stirling-like ideas. Also, Stirling used the Wallis product formula.
tsm [at] usna.edu
17. gowers Says:
That’s slightly embarrassing as I have the book but haven’t read that bit of it. But I would add to the previous comment that it is an excellent book in general, with beautiful discussions of bits of maths that often fall between the cracks of a typical undergraduate course. (For example, I was never taught the irrationality of pi and eventually learnt it from this book, where it formed part of a very interesting discussion of several related topics.)
18. Laurens Gunnarsen Says:
I can’t help thinking that this is an especially apposite moment to insert a ringing recommendation for Robert M. Young’s classic EXCURSIONS IN CALCULUS. Not only does the book include the very derivation of Stirling’s formula that Professor Gowers has presented here (on pp. 264-267), but it also offers several different approaches to deriving the deep and powerful Euler-Maclaurin summation formula, of which Stirling’s formula is a straightforward corollary.
In particular, on pp. 354-5 of his EXCURSIONS, Professor Young notes that summing the values of an analytic function f at the first n non-negative integers boils down to finding another function F related to it by
f(x) = F(x + 1) – F(x) = F'(x) + (1/2!)F”(x) + (1/3!)F”'(x) + … ,
since f(0) + f(1) + … + f(n-1) is then just F(n) – F(0). And now, to Euler, the striking formal resemblance between
f(x) = F'(x) + (1/2!)F”(x) + (1/3!)F”'(x) + …
and
y = x + ax^2 + bx^3 + …
naturally suggested solving for F in much the same way as one solves for x in terms of y in a power series of this sort. One simply writes
x = y + ry^2 + sy^3 + … ,
with r, s, … undetermined constants, and substitutes this form into the power series. Then one fixes r, s, … successively by requiring the coefficients of terms of like degree to match on both sides of the equation.
Of course there’s more to it than this in the end, but it’s hard to think of a more straightforward argument for putting
F'(x) = f(x) + rf'(x) + sf”(x) + … ,
and substituting back. And not only does this astonishingly simple idea actually work, it exposes the essence of the Euler-Maclaurin summation formula.
19. malkarouri Says:
The way I tend to remember this formula is by deriving it from equating the Poisson distribution with mean lambda by a Gaussian with mean and variance lambda. I first found this approach in David Mackay’s Information Theory, Inference, and Learning Algorithms. It is magic alright, but at least it was short enough for me to remember.
I wonder if one substitutes a Binomial distribution instead would one get C(2n, n)?
20. Dan Romik Says:
Dear Prof. Gowers,
Gil Kalai told me about your demystification of Stirling’s formula. Indeed, this easy result deserves to be better explained and understood, and I recall as an undergraduate I was quite frustrated at not being given a proper explanation for it. Then, about 10 years ago I discovered a new derivation of the constant sqrt(2 pi) which I believe to be slightly simpler than the one you presented, and conceptually better motivated, since it relates to the probabilistic viewpoint involving the central limit theorem. The difference with the approach you and a few of the readers mention is that I look at the cumulative probabilities of the symmetric binomial distribution instead of at just the “local” probabilities. By computing the sum of the first half of the binomial coefficients in a given row in two ways (first, using the obvious symmetry, and second, using a simple integration formula that converges to the integral of the Gaussian distribution), one gets the constant immediately.
My proof appeared in the American Math. Monthly 107 (2000), 556–557, and can be found on my web site. I wonder if you and others will think that my approach is indeed simpler than integrating powers of cosine – I find it hard to be objective about this, but of course both approaches are quite easy. By the way, I apologize for the pompous title of my paper (so pompous that I’m embarrassed to write it here) – in my defense, I was 21 when I wrote it…
21. gord Says:
Back to your question of why should nCr appear in integral of cos^n …
If you look at cos^n [or other functions that approach 1 smoothly & flatly from below…
and you keep reapplying them.. you get something that looks remarkably like a scaled
right half of a normal distribution… [seems to shrink left at 1/sqrt n ]…
My point is, you get the same thing with (1-x^2)^n or exp(-x*x/2) when raised to the n – the normal curve.
So to me, it seems like a product version of the central limit theorem – we keep reapplying by multiplying yet again by something roughly a normal distribution, and converges (nicely!) to the real normal distribution.
Turning the above sentence into math .. ie. stating why in the limit the repeated product converges to a scaled normal distribution… I havent done that yet – it might be obvious to others reading this blog …
Great, stimulating discussion/exposion, thanks!
[apologies for lack of latex]
A proof starting from the integral representation of gamma is given
in Creighton Buck’s “Advanced Calculus”, published over 50
years ago. This was likely not the first place it appeared. One
reason many people who derived Stirling’s formula used a
different argument was the desire for more terms in the
approximation to log gamma.
23. Neil Says:
Tim,
My favourite derivation of the constant is to look at the ratio of the binomial coefficients to the central binomial coefficient, use the fact that for small x,
(1+x) is very close to log(x), and obtain as a result the central limit theorem
for binomial coefficients directly. Then once you get that the coeffcients are
decaying about the middle like exp(-l^2/n), you compare to the integral of
exp(-x^2), and out pops the constant.
Of course, I each this in a generating functions course, and so I also show them
Euler Maclaurin as the trick to obtain everything but the value of the constant:
I actually use this example to motivate truly understanding the Euler MacM trick.
Neil
24. Takis Konstantopoulos Says:
I read the interesting discussion on the Stirling formula and
Tim Gowers’ method starts from the intuitive idea `replace sum by integral’.
It has the advantage that the crude Stirling approximation is obvious.
But the square root term (and the constant) are not obvious.
Besides, it does not give the link to the Central Limit Theorem,
which is the reason why $\sqrt{2\pi}$ appears.
See comments at the end of this, in reference to Laplace. He knew
that Stirling’s approximation is intimately connected with probabilities
of sums of a large number of random variables
and developed a general approximation method in his work on the analytical
theory of probabilities.
David Speyer’s method, starting from $n! = \int_0^\infty x^n e^{-x} dx$,
resembles very much Laplace’s method.
Also, as he observed, the connection to the CLT is apparent.
But there is one thing in David’s method,
which is not very natural.
Namely (following Laplace) it is better to make the
change of variable
$t=x/n$
first, instead of $t=x-n$. Then the gamma integral gives
$n! = n^{n+1} \int_0^\infty (t e^{-t})^n dt$
so that
$\frac{n!}{n^n e^{-n} \sqrt{n}} = \sqrt{n} e^n \int_0^\infty e^{-n g(t)} dt,$
where
$g(t) = t – \log t$
is a function with minimum at $t=1$,
and $g(1)=1$, $g'(1)=0$, $g”(1)=1$.
Following Laplace, the action is played by the second derivative of $g$,
so fixing and $\epsilon > 0$ pick $0< \delta < 1$ so that
$|g”(t) – 1| < \epsilon \mbox{ if } |t-1| 0, both bounds converge to zero. This means that \[ \frac{n!}{n^n e^{-n} \sqrt{n}} = o(1) + \sqrt{n} e^n \int_{1-\delta}^{1+\delta} e^{-n g(t)} dt.$
Now write
$g(t) = 1 + \frac{1}{2} g”(\tau)(t-1)^2$
for some $\tau=\tau(t)$ lying between $1$ and $t$, and therefore
$|g”(\tau) -1| 0$, (iv) the infimum of $g(x)$ over any compact
set not containing $x_0$ is strictly larger than $g(x_0)$,
(v) $f$ is continuous around $x_0$ with $f(x_0) \neq 0$.
Then
$I_n \sim f(x_0) e^{-(n+\frac{1}{2}) g(x_0)} \sqrt{2\pi/n g”(x_0)}.$
The proof is along the same lines as in the special case above.
Regarding the connection to the CLT:
Let $X_0, X_1, X_2$, … be a sequence of independent random variables,
all exponential with mean 1:
$P(X_j > t) = e^{-t}.$
Then $S_n = X_0+ X_1 + \cdots + X_n$ has density
$\frac{1}{n!} x^n e^{-x}.$
So the density of $(S_n-n)/\sqrt{n}$ is
$f_n(x) = \frac{1}{n!} (n+\sqrt{n}~x)^n e^{-n-\sqrt{n}~x} \sqrt{n}.$
A local version of the CLT tells us that, for all $x$ (in fact, uniformly),
$\lim_{n \to \infty} f_n(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$
Setting $x=0$ we obtain Stirling’s approximation.
Of course, the heavy-duty machinery invoked is not simpler
than the proof given above.
And here is an exact statement of the LOCAL CLT
(see, e.g. Fristedt and Gray, a Modern Approach to Probability Theory,
Thm 15.9, p.284):
Let $S_n = X_1+ … + X_n$ be sum of independent
and identically distributed random variables, where $X_1$ has
square integrable density, mean $\mu$, and finite variance $\sigma^2$.
Let $f_n$ be the density of $(S_n-n\mu)/\sqrt{n\sigma^2}$.
Then $f_n(x)$ converges (uniformly over all $x$)
to the standard Gaussian density.
25. Takis Konstantopoulos Says:
Hm… I had no intention in posting a comment with LaTeX code. I saw the earlier comments with nice formulae and thought that there was an automatic LaTeX converter embedded here. Wrong. My apologies.
http://www.ma.hw.ac.uk/~takis/commentonstirling.pdf
(If someone could tell me how to post properly, I’d appreciate!)
Sorry for the mess above!
26. Niall MacKay Says:
Your use of the piecewise-linear interpolation between the integers brings out the $O(n\log n)$, $O(n)$ and $O(\log n)$ terms so neatly that I wanted it very much to give the $O(1)$ term too — which it does, provided that
$\sum_{r=1}^\infty \{ (r+ {1\over 2})\log(1+{1\over r})-1\} = 1 - {1\over 2} \log(2\pi)$
which is true but which I cannot prove. Taylor-expanding the log and resumming in $\zeta(n)$, $n\geq 2$, doesn’t seem to help.
27. zeugirdorcar Says:
Nice. Thanks. I still think the $I_{n}$ integral looks like a rabbit from a hat…. I once saw Stirling’s approx following from Cauchy’s integral formula very naturally but I’ve been unable to reproduce it….. any way http://omega.albany.edu:8008/PascalT.pdf dots all the i’s for Stirling’s & the awsome thing is that with sage it was ALL DONE IN THE BROWSER!
|
2015-08-04 01:22:07
|
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|
http://dirkmittler.homeip.net/blog/archives/tag/error-message
|
## Kernel Update today, Downtime, Multiple Reboot-Attempts
Today, the PC which is hosting my site and blog, which I name ‘Phoenix’, received a kernel update.
Debian Team has not been following standard guidelines in their propagation of kernel updates, as the last 3 updates produced the same kernel-version number:
3.16.0-6-amd64
Because even Linux computers require a reboot after a kernel-update, this blog was temporarily off-line from about 13h05 until 13h25. I apologize for any inconvenience to my readers.
There is a fact about the build of Linux on this computer which I should bring up. I have the following on-board graphics-chip:
GeForce 6150SE nForce 430/integrated/SSE2
And this proprietary graphics driver is the only one, capable of working with the said graphics-chip:
NVIDIA 304.137
The graphics driver is installed from standard Debian repositories.
Somewhere between these software-packages there is a problem, which Debian Team has never been aware of, but which has existed ever since I installed Debian / Jessie on this computer. Directly after a reboot, the ability of the X-server to start, is not reliable. Sometimes, the X-server starts on the first try, but on other occasions I need to make 7 reboot attempts, before the X-server will start, and from one reboot-attempt to the next, I change nothing.
Once the X-server has started successfully, this graphics-chip will work 100% for 30 days !
I have been reluctant to point this out for the past few years, because if a Debian developer finds out about it, he will try to fix this problem. And when he does, he will brick my computer.
This afternoon, 7 reboots were in fact required, before the X-server started. That is why the reboot-procedure took 20 minutes of time.
(Updated 07/14/2018, 16h45 … )
## A First, Complicated Project at Circuit Design with NG-SPICE
One subject which I wrote about in an earlier posting, was that software exists by the name of ‘SPICE’, which stands for “Simulation Program with Integrated Circuit Emphasis”. There are several variants of this software in existence, but the version which I am focusing on for now is the Open-Source ‘NG-SPICE’ system, which needs to be bundled with numerous other packages under Linux, really to be useful. One important package is ‘ngspice-doc’, but there is a whole suite of Linux packages referred to as ‘gEDA’.
Simply having tested a few demo-projects, is not the same thing as actually having designed a circuit, and having witnessed that project ‘work’, at least according to the simulation. Just last night, I did the latter, in order to get a better, working grasp of how to use the software, and also, some idea of the sort of error messages and problems which invariably occur on a first-time basis. What this means is that I actually designed a circuit using the ‘gEDA Schemtic Editor’, which is also known as ‘gschem’, and then ran multiple simulations of the circuit, discovering at first that it had performance issues as I had imagined it, modifying it numerous times, and ending up with a version of the circuit, which I could be satisfied with for now.
The circuit which I was designing, actually involved MOSFETs, because those are the most important components in circuit-design today, and surely enough, I did run into initial problems. One of the tasks which we must complete, when using active components in SPICE, is to define the component, which is as fundamental as the fact that we also don’t just put a resistor, but must also specify what the Value of the resistor is in Ohms. Well with active components, we must do something similar, which also goes under the GUI heading of the Value attribute for the component. Therefore, MOSFETs, be they NMOS or PMOS, also have values, and by default, those values are defined by a Model Card, from which the computer can predict such physical properties about the NMOS or the PMOS transistor, as what its gate-capacitance is, how well it conducts when switched to conductive, conversely when the gate-voltage is zero, etc., etc., etc..
But, because NG-SPICE (v26) is advanced software – though still not the latest version – it may not require that the user defined all these parameters each time he or she considers designing a circuit, because standard component specifications exist.
By default, our MOSFETs have Reference Descriptors that begin with the letter “M”, and not with the letter “Q”, which would stand for a Bipolar Transistor, but which the GUI of ‘gschem’ suggests for the user when he first clicks a MOSFET into his circuit. So we override that, by editing the RefDes into a text-string that has the letter “M” followed without spaces by a number.
What I next proceeded to do, was to put MOSFET-transistors into my circuit, which from the GUI, only had 3 pins. This is a common way in which MOSFETs are often diagrammed, and looked something like this:
Believing that I could just accept what the GUI had constructed, I next tried to simulate the circuit, and received the error, which roughly stated “Unable to find Definition of Model.” This error-message wasted much of my time trying to solve, because I had in fact created a Model Card for the transistors which I was going to use, and at first, I despaired that NG-SPICE might not be as good as paid-for software. But I soon learned that indeed, the following example is a sufficient Model Card for an arbitrary NMOS transistor, with which circuits can be designed:
http://dirkmittler.homeip.net/text/NMOS1.mod.txt
Similarly, we can conjure a default PMOS transistor like so:
http://dirkmittler.homeip.net/text/PMOS1.mod.txt
In actual circuit-design, we’d drop the .TXT Filename-Extension, that makes the above examples readable in a Web-browser. Not only that, but we can also use the ‘gschem’ GUI, to embed such definitions directly into the Netlist, by giving them as a ‘Model’ attribute. So what was causing this error message, in my example? The fact is that MOSFETs are 4-pin components by nature. They have a hypothetical Source, a hypothetical Drain, a Substrate Electrode, and a Gate. It’s the voltage between the Gate and the Substrate Electrode, that finally determines how conductive the MOSFET is to become. By convention, many practical MOSFET-packages tie the Substrate Electrode together with the Source lead, which also happens to make the Source different from the Drain.
By telling NG-SPICE that we’re including a ‘MOSFET_TRANSISTOR‘ in our circuit, we’re telling this program to read 4 Nodes from the Netlist, to parse what the transistor is to be connected to. But, when the GUI only provides 3 arguments, an error ensues, that garbles the attempt of NG-SPICE to parse the Netlist. That’s all. Curiously, the reverse error does not happen. If I conjure a 4-lead MOSFET-symbol from the GUI, but specify a 3-lead MOSFET (more on that below), then I obtain a well-managed error message, that tells me what the problem is.
(Updated 06/14/2018 … )
Actually, the symbols above are also different in another way. In theory, one stands for an enhancement-mode, and the other, for a depletion-mode transistor. But, because under Linux, ‘this software is divided into two departments’, effectively, this does not matter.
The GUI allows schematics to be drawn in such a way, that Netlists result, while the actual NG-SPICE software emulates what these Netlists define. It’s in the emulation of the Netlists, that the decision is also made, as to whether a component is an enhancement-mode or a depletion-mode, or a subcircuit component… ( :1 )
(Updated 06/16/2018 : )
## Finding Out, How Many GPU Cores we have, Under Linux
One question which I see written about often on the Web, is how to find out certain stats about our GPU, under Linux. Under Windows, we had GUI-based programs such as ‘GPU-Z’, etc., but under Linux, the information can be just a bit harder to find.
I think that one tool which helps, is to have ‘OpenCL’ installed, as well as the command-line utility ‘clinfo’, which exists as one out of several packages, and as an actual, resulting command-name.
If we’re serious about programming our GPU, then having a GUI won’t help us much. We’d need to get dirty with code in that case, and then to have text-based solutions is suitable. But, if we’re just spectators in this sport, then two stats we may nevertheless want to know are:
1. How many GPU-Core-Groups do we have – since GPU-Cores are organized as Groups, and
2. How many actual Shader-Cores do we have in each Group?
Interestingly, the grouping of shader-cores, also represents how many vector-processors such GPU-computing tools as OpenCL see. And so, on the computer which I name ‘Klystron’, which is running Debian / Jessie, when typing in these commands as user, I get the following results:
dirk@Klystron:~$clinfo | grep units Max compute units: 4 Max compute units: 6 dirk@Klystron:~$ clinfo | grep multiple
Kernel Preferred work group size multiple: 1
Kernel Preferred work group size multiple: 64
dirk@Klystron:~\$
This needs some explaining. On ‘Klystron’, I have the proprietary, AMD packages for OpenCL installed, since that computer has both an AMD CPU and a Radeon GPU. And this means that the OpenCL version will be able to carry out computing on both. And so I have the stats for both.
In this case, the second entries reveal that I have 6×64 cores on the GPU.
## Freshly switched to KDE 4 or Plasma 5, and unable to Browse Network Shares using Dolphin?
I just installed Plasma 5 from the package-manager, on my tower-computer named ‘Plato’, only to find that for some time, I was unable to browse Windows File Shares – i.e. ‘SMB Shares’ – casually, just using the ‘Dolphin’ File Manager. Yet, I was able to mount these shares using ‘Smb4K’, making them visible in my local folders as though there.
Dolphin was showing me an essentially empty set of icons when displaying the Network.
As it turns out, we need to install a package named:
‘kio-extras’
Which will give Dolphin the additional plug-ins it needs, to recognize ‘smb://’ URIs. If our Plasma 5 desktop manager was set up professionally, then the person doing so would know about such details. But when individuals set up KDE or Plasma for the first time, we need to learn such details first-hand.
As an added note, we might find that when we click on the Trash widget in our Panel, which I left just at the right-most end, we may get the error-message to the effect that ‘trash:/’ was a corrupted URL. Yet, from within Dolphin, the trash bin displays just fine.
In my case this was happening, because I did not have Dolphin set up as my default File Management application, in my Plasma 5 Settings, where instead I had an application selected which would have been appropriate to an LXDE desktop, and which does not recognize URIs that begin with ‘trash:/’. Switching this setting to make Dolphin my Default File Manager, fixed this problem.
Dirk
|
2019-12-12 01:26:04
|
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|
http://assert.pub/arxiv/nlin/nlin.si/
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### Top 10 Arxiv Papers Today in Exactly Solvable And Integrable Systems
##### #1. Large-order asymptotics for multiple-pole solitons of the focusing nonlinear Schrödinger equation
###### Deniz Bilman, Robert Buckingham
We analyze the large-$n$ behavior of soliton solutions of the integrable focusing nonlinear Schr\"odinger equation with associated spectral data consisting of a single pair of conjugate poles of order $2n$. Starting from the zero background, we generate multiple-pole solitons by $n$-fold application of Darboux transformations. The resulting functions are encoded in a Riemann-Hilbert problem using the robust inverse-scattering transform method recently introduced by Bilman and Miller. For moderate values of $n$ we solve the Riemann-Hilbert problem exactly. With appropriate scaling, the resulting plots of exact solutions reveal semiclassical-type behavior, including regions with high-frequency modulated waves and quiescent regions. We compute the boundary of the quiescent regions exactly and use the nonlinear steepest-descent method to prove the asymptotic limit of the solitons is zero in these regions. Finally, we study the behavior of the solitons in a scaled neighborhood of the central peak with amplitude proportional to $n$. We...
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Authors: 2
Total Words: 16855
Unqiue Words: 2913
##### #2. Constructing Initial Value Spaces of Lattice Equations
###### Nalini Joshi, Sarah Lobb, Matthew Nolan
In this paper, we examine the initial values for integrable lattice equations, in particular, the lattice equations classified by Adler et al (2003), known as ABS equations. We compactify and resolve singularities explicitly to construct initial value spaces and show how the resulting initial-value space of an ABS equation reduces to that for a discrete Painlev\'e equation. The construction also leads us to the discovery of new reductions of lattice equations that are based on the coordinates that define exceptional planes in the space of initial values for that equation.
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Sample Sizes : None.
Authors: 3
Total Words: 5817
Unqiue Words: 1398
##### #3. On The Painleve Property For A Class Of Quasilinear Partial Differential Equations
###### Stanislav Sobolevsky
The last decades saw growing interest across multiple disciplines in nonlinear phenomena described by partial differential equations (PDE). Integrability of such equations is tightly related with the Painleve property - solutions being free from moveable critical singularities. The problem of Painleve classification of ordinary and partial nonlinear differential equations lasting since the end of XIX century saw significant advances for the equation of lower (mainly up to fourth with rare exceptions) order, however not that much for the equations of higher orders. Recent works of the author have completed the Painleve classification for several broad classes of ordinary differential equations of arbitrary order, advancing the methodology of the Panleve analysis. This paper transfers one of those results on a broad class of nonlinear partial differential equations - quasilinear equations of an arbitrary order three or higher, algebraic in the dependent variable and including only the highest order derivatives of it. Being a first...
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###### Tweets
SciencePapers: On The Painleve Property For A Class Of Quasilinear Partial Differential Equations. https://t.co/zpPrvqOTAJ
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Authors: 1
Total Words: 2113
Unqiue Words: 840
##### #4. On a new integrable generalization of the Toda lattice and a discrete Yajima-Oikawa system
###### Takayuki Tsuchida
We propose a new integrable generalization of the Toda lattice wherein the original Flaschka-Manakov variables are coupled to newly introduced dependent variables; the general case wherein the additional dependent variables are vector-valued is considered. This generalization admits a Lax pair based on an extension of the Jacobi operator, an infinite number of conservation laws and, in a special case, a simple Hamiltonian structure. In fact, the second flow of this generalized Toda hierarchy reduces to the usual Toda lattice when the additional dependent variables vanish; the first flow of the hierarchy reduces to a long wave-short wave interaction model, known as the Yajima-Oikawa system, in a suitable continuous limit. This integrable discretization of the Yajima-Oikawa system is essentially different from the discrete Yajima-Oikawa system proposed in arXiv:1509.06996 and studied in arXiv:1804.10224. Two integrable discretizations of the nonlinear Schr\"odinger hierarchy, the Ablowitz-Ladik hierarchy and the...
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##### #5. Non-abelizations of quadratic ODE-systems
###### V. Sokolov, T. Wolf
New non-abelizations of integrable quadratic homogeneous systems of ODEs with two dependent variables are found.
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Authors: 2
Total Words: 4501
Unqiue Words: 1026
##### #6. Algebraic Properties of Quasilinear Two-Dimensional Lattices connected with integrability
###### I. T. Habibullin, M. N Kuznetsova
In the article a classification method for nonlinear integrable equations with three independent variables is discussed based on the notion of the integrable reductions. We call the equation integrable if it admits a large class of reductions being Darboux integrable systems of hyperbolic type equations with two independent variables. The most natural and convenient object to be studied within the frame of this scheme is the class of two dimensional lattices generalizing the well-known Toda lattice. In the present article we deal with the quasilinear lattices of the form $u_{n,xy}=\alpha(u_{n+1} ,u_n,u_{n-1} )u_{n,x}u_{n,y} + \beta(u_{n+1},u_n,u_{n-1})u_{n,x}+\gamma(u_{n+1} ,u_n,u_{n-1} )u_{n,y}+\delta(u_{n+1} ,u_n,u_{n-1})$. We specify the coefficients of the lattice assuming that there exist cutting off conditions which reduce the lattice to a Darboux integrable hyperbolic type system of the arbitrarily high order. Under some extra assumption of nondegeneracy we described the class of the lattices integrable in the sense...
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Sample Sizes : None.
Authors: 2
Total Words: 10663
Unqiue Words: 2077
##### #7. The resonant structure of Kink-Solitons in the Modified KP Equation
###### Jen-Hsu Chang
Using the Wronskian representation of $\tau$-function, one can investigate the resonant structure of kink-soliton and line-soliton of the modified KP equation. It is found that the resonant structure of the the soliton graph is obtained by superimposition of the two corresponding soliton graphs of the two Le-Diagrams given an irreducible Schubert cell in a totally non-negative Grassmannian $Gr(N,M)_{ \geq 0}$. Several examples are given.
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##### #8. Integrable discrete autonomous quad-equations admitting, as generalized symmetries, known five-point differential-difference equations
###### R. N. Garifullin, G. Gubbiotti, R. I. Yamilov
In this paper we construct the autonomous quad-equations which admit as symmetries the five-point differential-difference equations belonging to known lists found by Garifullin, Yamilov and Levi. The obtained equations are classified up to autonomous point transformations and some simple non-autonomous transformations. We discuss our results in the framework of the known literature. There are among them a few new examples of both sine-Gordon and Liouville type equations.
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##### #9. The Kowalewski's top revisited
###### Franco Magri
The paper is a commentary of one section of the celebrated paper by Sophie Kowalewski on the motion of a rigid body with a fixed point. Its purpose is to show that the results of Kowalewski may be recovered by using the separability conditions obtained by Tullio Levi Civita in 1904.
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##### #10. Sato Grassmannian and Degenerate Sigma Function
###### Julia Bernatska, Victor Enolski, Atsushi Nakayashiki
The degeneration of the hyperelliptic sigma function is studied. We use the Sato Grassmannian for this purpose. A simple decomposition of a rational function gives a decomposition of Pl\"ucker coordinates of a frame of the Sato Grassmannian. It then gives a decomposition of the tau function corresponding to the degeneration of a hyperelliptic curve of genus $g$ in terms of the tau functions corresponding to a hyperelliptic curve of genus $g-1$. Since the tau functions are described by sigma functions, we get the corresponding formula for the degenerate hyperelliptic sigma function.
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2019-01-21 19:34:45
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{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.71547532081604, "perplexity": 2353.3266310912495}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583807724.75/warc/CC-MAIN-20190121193154-20190121215154-00310.warc.gz"}
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http://www.maplesoft.com/support/help/MapleSim/view.aspx?path=Task/PlotFunctionAndDerivatives
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Plot a Function and Its Derivatives - Maple Help
Home : Support : Online Help : Tasks : Calculus - Differential : Derivatives : Task/PlotFunctionAndDerivatives
Plot a Function and Its Derivatives
Description Plot a function and its derivatives in a specified interval.
Graph of $f,f\prime ,$and $f″$in a Specified Interval Enter the function $f\left(x\right)$ to be studied, and the interval on which to plot it. $f\left(x\right)=$ Interval: [ , ]
Commands Used
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2016-05-24 08:01:13
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https://crypto.stackexchange.com/questions/9054/a-block-cipher-with-independent-keys-for-each-round/9055
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# A block cipher with independent keys for each round
I want to modify a standard block cipher in the following way. I replace each round key by a key picked at random. Is this block cipher as secure as the original one ?
Thank you.
EDIT
Some missing information related to the question, according to comment by OP to one of the answers:
My question is related to the key-schedule of AES, I wonder if we can replace it by something more stringent and be sure that the security will not be not worst than the original one.
• One of my experiments involves using Keccak (c=1024) as the key schedule, with the squeeze truncated to the block size (128-bits for AES), and a new squeeze for each round key. This should maintain nonlinearity of round keys, and allow a more appropriately sized cipher key length. Key changes are substantially more expensive. – Richie Frame Nov 4 '13 at 3:56
• I would consider the practical problem with such a large key algorithm to be "where to get independent bits as key material". As the keying material is likely not going to be independent, related key attack may be concern. – user4982 Jan 31 '14 at 17:40
John Kelsey, Bruce Schneier and David Wagner proposed paper "Key-Schedule Cryptanalysis of IDEA, G-DES, GOST, SAFER, and Triple-DES" and they presented new attacks on key schedules of the block ciphers. About "A 768-bit DES variant uses independent round subkeys" they said:
A 768-bit DES variant uses independent round subkeys [Ber83]. This variant will be much weaker in some situations: there is a very simple related-key attack needing just 15 related keys and 60 chosen plaintexts. Obtain the encryptions $E(k, p)$ and $E(k' , p)$, where k is obtained from k by flipping some bits in the last round subkey; this can be thought of as a differential $1R$ attack with a characteristic of probability 1. The last round subkey can be recovered with four chosen plaintexts, and then we can peel off the last round and repeat the attack on 15-round DES. This attack can also be optimized for the case when related key queries are very expensive to achieve a complexity of one related key and $2^{16}$ or so chosen plaintexts. For nearly any product block cipher, if it’s possible to flip bits in a cipher’s expanded key, it’s possible to mount an XOR differential attack on the last round of the cipher. This may be useful in attacking some systems that leave expanded keys vulnerable to change.
So if the algorithm is DES the answer is: NO.
Charles Bouillaguet, Patrick Derbez, Orr Dunkelman, Nathan Keller and Pierre-Alain Fouque proposed "Low Data Complexity Attacks on AES". They said:
We present several attacks on up to four rounds of AES requiring up to ten chosen plaintexts. Most of the attacks are based on the meet-in-the-middle approach. Some of the attacks exploit heavily the AES key schedule, while others apply even if the subkeys in AES are replaced by independent subkeys. The attacks are summarized in Table 1.
The attacks are applicable to up to 7-rounds of Rijndael. Since the minimal number of rounds in the Rijndael parameter settings proposed for AES is 10, these attacks does not endanger the security of the cipher.
By now there isn't any attacks against full-rounds of Rijndael key schedule or full-rounds of Rijndael with independent subkeys.
So if the algorithm is AES the answer is: By now YES
But the answer is depending on future researches. And in general the answer is depending on the algorithm.
• Actually, the attack that Kelsey et al proposes for DES with independent subkeys can easily be applied to AES with independent subkeys. Hence, if we consider related key attacks, the answer for AES would be No as well. – poncho Nov 2 '13 at 10:34
• In the related key scenario, allowing independent round keys always offers a lot more variables to modify, but the attacker could also just use the "valid" combinations, which agree with the usual key schedule. The attack with independent round keys therefore is stronger than the common related key attack (probably even strictly stronger). And therefore "security is weaker". – tylo Sep 7 '15 at 11:22
This question can be answered in several way depending on the exact meaning you intend for more secure.
First answer: No, it is not more secure in general. The most you can expect is "at least as secure" not "more secure". A typical example of this behavior is Even-Mansour encryption where using twice the same key is as secure as using two independent random keys (see http://eprint.iacr.org/2011/541.pdf).
Second answer: Yes, it is at least as secure, because the goal of the key schedule is usually to derive random looking round keys from the cipher key.
Third answer: No, it is not as secure, because it is more vulnerable to related key attacks. Or more precisely, related round-keys attacks become ordinary related key attacks which weakens the security level.
Fourth answer: No, it is not secure, because it is very unlikely that the modified cipher attain the expected security bound for a cipher with such a long key. [For example, using this idea with AES gives a 1000+-bit key and I would be surprised to hear that you attain the $2^{1000}$ security level.] This is one of the reasons for which Triple-DES with 2 keys is usually prefered to Triple-DES with 3 keys.
So it would be useful to specify more precisely why you would like to do this and what is the definition of security you are considering in your context.
• In your fourth answer, you want to says "not more secure?". I'm not looking for a $2^{1000}$ security level. I wonder if this modified AES (with fixed random round keys) is better able to achieve the security that the original AES should fulfil (attain). – Dingo13 Jul 5 '13 at 21:33
• No. I want to say not secure. Because many theoretical cryptographers consider that something that does not require close to exhaustive key search is broken. I just wanted to emphazise that you need to define the security notion you have in mind more precisely. – minar Jul 5 '13 at 21:44
• I don't have a specific security notion in mind. I just wonder if we had to choose between an AES with a key schedule and an AES with independent round keys, which one should be preferable ? From the point of view of security, not key material length... In other words which one seems to be the most vulnerable... – Dingo13 Jul 5 '13 at 21:52
• Then consider that I don't want more security than AES should give me... But I want to lessen the power of well known attacks. – Dingo13 Jul 5 '13 at 22:00
• What kind of attacks ? If you only consider crypto attacks on the block cipher in the standard model and isolated from its context, your proposal is normally fine. However, you are weakening the security against related-key attacks and probably easy the work of an adversary that tries to attack a combination of key-exchange + block cipher. So if your goal is just to increase security, it is preferable not to do that. – minar Jul 6 '13 at 7:33
If a cipher has independent round keys, then it is trivially susceptible to a meet-in-the-middle attack. Independence of the rounds means you can brute force the first half of the rounds and second half of the rounds separately. A cipher with independent round keys totaling $n$ bits can be brute forced in $2^{n/2}$ time. So, from the point of view considering key length alone, this is not a very good cipher.
However, it is likely that $n/2$ is much greater than the key length of the thing you started with. And it seems very plausible that the cipher could actually achieve security in the neighborhood of $2^{n/2}$ (as opposed to $2^{n/2}$ just being the obvious upper bound). Heuristically at least, the key schedule is supposed to derive many "independent-looking" keys from a single key. So the cipher may have been designed and analyzed based on the heuristic that is now the reality.
So I don't think it's a big stretch to consider such a cipher to be secure with $n/2$ bits of security, though it's not the most aesthetically pleasing way to get $n/2$ bits of security.
• "So, from the point of view considering keylength alone, this is not a very good cipher." I'm agree with you that it's totally useless to desire such a long key. Meet-in-the middle-attack is not feasible (not interesting) with such a long key. In fact I don't want $n/2$ bits of security (with $n=10 \times 128$), my question is just about the relative security. I wonder if we can have the same security than the standard AES by having independent keys by round, that's all. – Dingo13 Jul 13 '13 at 14:32
No, it's less secure because:
• you're not using the algorithm in the way the author(s) designed it
• it hasn't been subjected to scrutiny by trained cryptographers
If you don't know the above already, you certainly don't have enough experience in cryptography to tinker with the inner-workings of algorithms/modes. Stick with the standard algorithm/mode, it's a much safer bet.
• However the key is longer in the proposed modified scheme. Besides, if we consider the original scheme, the round keys are related from the master key, intuitively should this not imply a lower security ? – Dingo13 Jul 5 '13 at 17:39
• "No, it's less secure because: - you're not using the algorithm in the way the author(s) designed it -it hasn't been subjected to scrutiny by trained cryptographers" Authors have designed a system with related (derivable) round keys by a desire of short key length ? Not ? – Dingo13 Jul 5 '13 at 17:44
• The key being longer does not inherently discount the possibility of having introduced a weakness by changing the algorithm. Besides — if you're dealing with a modern cipher of at least 128 bits, key length is hardly a concern. At 256 bits, it would likely never be brute-forced before the heat death of the universe. So what problem are you trying to solve, exactly? – Stephen Touset Jul 5 '13 at 18:48
• Concrete example: There are better attacks on AES-256/192 than AES-128 because the key schedule is weaker. – pg1989 Jul 5 '13 at 20:18
• @pg1989 Which only apply if you use AES with non random keys. Related key attacks are only relevant if you want to use AES in unusual ways, for example by building a hash function from it. – CodesInChaos Jul 5 '13 at 20:28
This is at least as secure as the original cipher.
The only case I can think of where it would be less secure is if the security of the cipher relied on some special relation between the round keys, but I don't know of any ciphers that have this requirement. Most ciphers derive their round keys from the encryption key in a linear way.
One example of a cipher that does something similar to this is Blowfish. It uses the encryption function as a pesudo-random function to generate the round keys.
• Thank you. This was what I think but what about key-related attacks ? – Dingo13 Jul 5 '13 at 18:38
• I would argue that in a cipher where key size is equal to output block size, no security is gained. Assuming the cipher is capable of generating all (or even most) bit sequences of its block size, encryption as described would be effectively no different than encryption with some key of the standardized length. – Stephen Touset Jul 5 '13 at 18:50
• In the random case, there is no relation between the round keys. This makes the cipher more difficult to attack. For example in AES, the key schedule is invertible, so if you have one round key, you have them all. This is not the case with random round keys. – user7487 Jul 5 '13 at 18:58
• @Stephen Touset. I've not well understood what you are saying. You say that I can introduce a weakness by replacing round keys by independent random keys ? – Dingo13 Jul 5 '13 at 19:45
• @StephenTouset, there are way more than $2^{256}$ possible 256-bit permutations. – otus Aug 4 '15 at 6:17
A good block cipher makes each bit of the ciphertext depend on every bit of the key and every bit of the plaintext. By making the key longer and only parts of it involved in each round you break that property for algorithms designed to achieve it. It is therefore less secure.
• You speak about the security that we should have corresponding to the bitlength of the key. I don't want security equal to the bitlength of the key, I wonder if for a security of 128 bits, having one key by round is as secure than the original AES. – Dingo13 Jul 13 '13 at 14:22
• You don't interpret correctly my question. – Dingo13 Jul 13 '13 at 14:23
• Answered as asked, but after the edit, I recommend the answer from ir01. – Árni St. Sigurðsson Jul 14 '13 at 14:34
Short answer:Yes, as long as you don't misuse it, it will be just as secure as the original block cipher. Much like a live grenade if you aren't careful it will explode in your face.
DISCLAMER: It's good to be curious about cryptographic primitives but this stuff will blow up in your face unexpectedly. I've written code vulnerable to hash length extension attacks. Don't use custom stuff in production.
The goal of a block cipher's key schedule is to produce somewhat random round keys. This is a balance between performance when frequent re-keying is required (EX:webserver with 1000 clients) and mitigating against programmer stupidity(more later), AES has a very simple key schedule and especially 192 and 256 bit variants are showing strain.
AES128 can be turned into AES(128*10) by getting rid of the key schedule and feeding in round keys directly. Doing so is extremely dangerous but only if you misuse the resulting block cipher (much like you can misuse AES itself by using it in ECB mode). The danger here comes from related key attacks. If the attacker can flip key bits individually (which in this case translate directly to round key bits) they can find the key easily. This is like reusing a nonce in ECDSA. But this is only possible if you give them related keys. Like if you put a message number counter in the key. That's not some far-fetched rare thing. Remember WEP? The utterly broken standard for WIFI network security. It called for putting a message counter in the RC4 key. LAter on someone found a related key attack. Thanks to its key schedule, unmodified AES can cope with that. AES(128*10) will, metaphorically, explode.
AES(128*10) round keys must be generated using a strong pseudorandom function (like KACCAK as you were doing before). AES itself though, is such a function so if code size is a concern don't hesitate to use normal AES in CTR mode to generate round keys for your custom AES1280. The result will be no less secure than the original.
One more disclaimer: AES is tricky to implement securely. Start with an existing, known safe implementation like the one in openssl.
• "The danger here comes from related key attacks." How so? Randomly chosen round keys are LESS related than the ones generated by the standard key schedule algorithm. The attackers ability to flip (round) key bits assumes that the attacker has any influence on how the keys are generated or on the process of execution of the algorithm. If we assume the latter then side-channel attacks must be considered. DFA attack on a standard implementation will recover the round keys be they from the standard key schedule or random (the random case requiring repeating the attack for each round). – astraujums Feb 7 '18 at 15:05
If each round key is of the same size of the block to be encrypted and all round keys are truly randomly generated independent from each other, then we have OTP. This is similar to using a truly randomly generated key of the same size of the plaintext to be encrypted. Each bit of this key is used only one-time to encrypt one plaintext bit.
• AES key is already at least as long as the block, but that does not make it equivalent to OTP (among other things because it's typically used to encrypt many blocks). Why would independent round keys to that? – otus Aug 4 '15 at 6:12
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2020-09-30 14:34:05
|
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|
https://mathematica.stackexchange.com/questions/111124/the-numerical-solution-of-a-nonlinear-ode-boundary-value-problem
|
# The numerical solution of a nonlinear ODE: boundary value problem
I want to solve the following nonlinear ordinary differential equation with boundary values $f(0)=1$ and $f(1)=1$:
$${\frac {{\rm d}^{2}}{{\rm d}{x}^{2}}}f \left( x \right) ={\dfrac { \left( 51-51\,x \right) \left( 1+ \left( {\dfrac {\rm d}{ {\rm d}x}}f \left( x \right) \right) ^{2} \right) ^{3/2}}{25}}+{ \frac {1+\left({\dfrac{\rm d}{{\rm d}x}}f \left( x \right)\right)^2 }{f \left( x \right) }}$$
But:
NDSolve[{y''[x] ==
204/100 (1 - x) (1 + y'[x]^2)^(3/2) + (1 + y'[x]^2)/y[x],
y[0] == 1, y[1] == 1}, y[x], {x, 0, 1}]
gives only the following message:
Power::infy: Infinite expression 1/0. encountered. >>
Infinity::indet: Indeterminate expression 0. ComplexInfinity encountered. >>
NDSolve::ndnum: Encountered non-numerical value for a derivative at x == 0.. >>
Is it because something is wrong with the ODE itself? or I will have to use some special options in NDSolve in order to handle such an ODE?
• Could you share a little about the physics that this equation is solving? That way someone may be able to tell you about any well established boundary condition issue or specification with physical insight, perhaps? It would seem that it is the final term in your equation (1 + Derivative[1][y][x]^2)/y[x] causing the issue. Mar 26, 2016 at 12:49
• It's from a Chinese article on determining the curve shape of a one dimensional liquid surface with specific surface tension and pressure difference at the surface. So you will see something from the nonlinear ODE that is similar to and actually derived from the curvature radius of the liquid profile curve. Mar 28, 2016 at 3:11
• Could you please cite the paper by any chance? Thank you. Mar 28, 2016 at 12:50
EDIT #2 (12.05.16)
I am greateful to the author of the OP for pointing out an error I made in EDIT #1 (which was also made by himself and user drN).
This leads me to revise my revision (EDIT #1) and return to the statement of the original solution: There is no solution to the original problem.
But can we can some words more on the question why there is no solution and "how far away from solubility" is the original system.
Experimenting as suggested in the first solution we find that the second derivative is so strong that it bends the curve up across the line y = 1 already for x < 1.
Hence we can possibly reach the goal y(x=1)=1 by reducing the impact of the second derivative by changing the formulas.
There are, of course, many ways to accomplish this.
We consider two possibilities
1) Reduce impact of second derivative by reducing the power of (1+y'[x]^2)
We already did the replacement 3/2 -> 1/2 "unnoticed" in EDIT #1. So these results can be reinterpreted in the current framework and are still of interest.
Letting 3/2->p we find for a critical power pc = 0.5917 defined such that there is no soluton for p > pc
With[{p = 0.5917},
yy[x_] = y[x] /.
NDSolve[{y''[x] ==
204/100 (1 - x) (1 + y'[x]^2)^p + (1 + y'[x]^2)/y[x],
y[0] == 1, y[1] == 1}, y[x], {x, 0, 1.5}][[1]]];
Plot[{1, yy[x]}, {x, 0, 1}] (* graph not shown here *)
{yy[0], yy[1]}
{1., 1.}
2) Reduce impact of second derivative by reducing the multiplicative factor "a" instead of 204/100
By trial and error we find a critical value ac = 1.3107 defined such that for a>ac there is no solution possible.
With[{a = 1.3107},
yy[x_] = y[x] /.
NDSolve[{y''[x] ==(*204/100 *)
1.3107 (1 - x) (1 + y'[x]^2)^(3/2) + (1 + y'[x]^2)/y[x], y[0] == 1,
y[1] == 1}, y[x], {x, 0, 1.5}][[1]]];
Plot[{1, yy[x]}, {x, 0, 1}] (* graph not shown here *)
{yy[0], yy[1]}
{1., 1.}
EDIT #1 (26.03.16)
Let us apply the shooting method, i.e. solve the equation with initial conditions and look for the value of y[1].
Let y[0] = 1 and y'[0] equal to some value v which we shall vary in order to hopefully get y[1] = 1:
Revising my previous statement we find that there are even two different solutons to the problem, i.e. the solution is not defined uniquely by the boundary conditions. This meakes the problem even more interesting.
I was mistaken in thinking of a monotoneous dependence of y[1] of y'[0] (=v). Closer inspection reveals that this was wrong.
Here's the code to experiment with:
v = -10; yy[x_] =
y[x] /. NDSolve[{y''[x] ==
204/100 (1 - x) (1 + y'[x]^2)^(3/2) + (1 + y'[x]^2)/y[x], y[0] == 1,
y'[0] == v}, y[x], {x, 0, 2}][[1]];
Plot[{1, yy[x]}, {x, 0, 2}]
For a systematic approach let uns define the value y1 = y[1] as a function of the slope v at x = 0
y1[v_] := (y[x] /.
NDSolve[{y''[x] ==
204/100 (1 - x) (1 + y'[x]^2)^(1/2) + (1 + y'[x]^2)/y[x], y[0] == 1,
y'[0] == v}, y[x], {x, 0, 2}][[1]]) /. x -> 1
The plot shows that the two values v ~= -3.5 and v ~= -2.6 make y[1] == 1:
Plot[y1[v] - 1, {v, -4, -2}, AxesLabel -> {"y'[0]", "y[1]-1"}]
Greater precision is obtained manually adjusting v such that y1 = 1.
For solution 1
v = -2.582496;
yy1[x_] = y[x] /.
NDSolve[{y''[x] ==
204/100 (1 - x) (1 + y'[x]^2)^(1/2) + (1 + y'[x]^2)/y[x], y[0] == 1,
y'[0] == v}, y[x], {x, 0, 2}][[1]];
yy1[1]
(* Out[491]= 0.99999 *)
For solution 2:
v = -3.59259;
yy2[x_] = y[x] /.
NDSolve[{y''[x] ==
204/100 (1 - x) (1 + y'[x]^2)^(1/2) + (1 + y'[x]^2)/y[x], y[0] == 1,
y'[0] == v}, y[x], {x, 0, 2}][[1]];
yy2[1]
(* Out[496]= 1. *)
Plotting both solutons together
Plot[{1, yy1[x], yy2[x]}, {x, 0, 1.2},
PlotLabel -> "The two solutons of the boundary value problem",
AxesLabel -> {"x", "y[x]"}, PlotRange -> {0, 1.5}]
Original solution
Thus boundary value problem has no soluton.
Here we go to show it:
Let us solve the equation with initial conditions and look for the value of y1 obtained.
Let y[0] = 1 and y'[0] equal to some value v which we shall vary in order to hopefully get y1 == 1:
Here's the code to experiment with:
v = -10; yy[x_] =
y[x] /. NDSolve[{y''[x] ==
204/100 (1 - x) (1 + y'[x]^2)^(3/2) + (1 + y'[x]^2)/y[x], y[0] == 1,
y'[0] == v}, y[x], {x, 0, 2}][[1]];
Plot[{1, yy[x]}, {x, 0, 2}]
• Could this be a version related issue? I just started a fresh kernel and the problem solved fine "right out of the box" using NDSolveValue: NDSolveValue[{(y^\[Prime]\[Prime])[x] == 204/100 (1 - x) (1 + Derivative[1][y][x]^2)^(1/2 ) + ( 1 + Derivative[1][y][x]^2)/y[x], y[0] == 1, y[1] == 1}, y[x], {x, 0, 1}] Mar 26, 2016 at 13:09
• @drN You wrote 1/2 instead of 3/2. In this case it works out fine also in my version which is 10.1.0 for Microsoft Windows (64-bit) (March 24, 2015). But the problem was a different one. Mar 26, 2016 at 13:32
• Thank you! I found by using Newton's method it is possible to find the two initial conditions more accurate, with different start points: (1)FindRoot[obj[x] == 1, {x, -2}, Evaluated -> False] // NumberForm[#, 15] & (1) FindRoot[obj[x] == 1, {x, -4}, Evaluated -> False] // NumberForm[#, 15] &`. The two start points were of course easily be identified by plotting the curve as in the answer. thanks Mar 28, 2016 at 3:22
• @user6043040 I am not aware of the literature on the topic, but I would start with en.wikipedia.org/wiki/Shooting_method Mar 28, 2016 at 8:03
• @user6043040 You are right, I changed the exponent to 1/2 because it was done in the comments of drN and you and I didn't check it. I shall make a brief EDIT in my answer to summaize the status. May 12, 2016 at 14:23
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2022-05-27 22:13:40
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http://www.varsitytutors.com/act_math-help/how-to-find-the-solution-of-a-rational-equation-with-a-binomial-denominator
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# ACT Math : How to find the solution of a rational equation with a binomial denominator
## Example Questions
### Example Question #1 : How To Find The Solution Of A Rational Equation With A Binomial Denominator
For the equation , what is(are) the solution(s) for ?
Explanation:
, can be factored to (x -7)(x-3) = 0. Therefore, x-7 = 0 and x-3 = 0. Solving for x in both cases, gives 7 and 3.
### Example Question #2 : How To Find The Solution Of A Rational Equation With A Binomial Denominator
Simplify:
Explanation:
In order to begin this kind of a problem, it's key to look at parts of the rational expression that can be simplified.
In this case, the denominator is an already-simplified binomial; however, the numerator can be factored.
The roots will be numbers that sum up to but have the product of .
The options include:
When these options are summed up:
We can negate the last three options because the first option of and fulfill the requirements. Therefore, the numerator can be factored into the following:
Because the quantity appears in the denominator, this can be "canceled out." This leaves the final answer to be the quantity .
### Example Question #3 : How To Find The Solution Of A Rational Equation With A Binomial Denominator
Simplify:
Explanation:
In order to begin this kind of a problem, it's key to look at parts of the rational expression that can be simplified.
In this case, the denominator is an already-simplified binomial; however, the numerator can be factored through "factoring by grouping." This can be a helpful idea to keep in mind when you come across a polynomial with four terms and simplifying is involved.
can be simplified first by removing the common factor of from the first two terms and the common factor of from the last two terms:
This leaves two terms that are identical and their coefficients, which can be combined into another term to complete the factoring:
Consider the denominator; the quantity appears, so the in the numerator and in the denominator can be cancelled out. The simplified expression is then left as .
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2016-10-25 13:56:44
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http://web.emn.fr/x-info/sdemasse/gccat/Knon-deterministic%20automaton.html
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3.7.157. Non-deterministic automaton
A constraint for which the catalogue provides a non -deterministic automaton without counters and without array of counters. For the mentioned constraints it turn out that non -determinism is due to the fact that we introduce transitions labelled by the potential values of a counting variable to a single final state (i.e., see Figures 5.16.4, 5.49.4, and 5.298.4).
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2019-04-23 20:29:36
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https://studydaddy.com/question/his-103-week-1-quiz
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QUESTION
# HIS 103 Week 1 Quiz
This archive file of HIS 103 Week 1 Quiz shows the solutions to the following problems:
1. The fusion of Aryan traditions and Dravidian beliefs and values laid the foundation for what religion?2. All of the following helped the Achaemenids maintain their empire except:3. What Zhou political theory stated that heavenly powers granted the right to govern?4. The earliest known cities developed between what two rivers, also known as the cradle of civilization?5. Which of the following groups did not rule over Egypt?6. Civilization in China developed around these two rivers:7. Hammurabi codes did all but:8. This 6th century BCE ruler extended his empire from the Indus River to the Aegean Sea and from Armenia to the First Cataract of the Nile.9. What society thrived from 3000 BCE-1900BCE around the Indus River?10. Which characterized Mesopotamia:
• $7.79 ANSWER Tutor has posted answer for$7.79. See answer's preview
*** *** Week 1 ****
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2017-09-25 11:30:12
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https://www.mersenneforum.org/showthread.php?s=8f6c9270e8f52ab97ae2ee6664fe8cd5&t=20743&page=2
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mersenneforum.org Sieving for CRUS
User Name Remember Me? Password
Register FAQ Search Today's Posts Mark Forums Read
2015-12-13, 05:52 #12
VBCurtis
"Curtis"
Feb 2005
Riverside, CA
27×3×13 Posts
Quote:
Originally Posted by LaurV S428 200k
The k's remaining are low-weight enough that it's pretty likely the entire file will be tested; so this question is roughly equivalent to "what sieve depth minimizes the time to sieve plus the time to LLR the file?" In the short run, it's about finding the point where breaking off a piece (say, 200-300k) for LLR followed by more sieving on 300k-1M is better than just sieving more on the 200k-1M file.
If you test the 300k-1M sieve speed vs 200k-1M, you'll find that there's very little speed gain, meaning the 200-300 piece is, in a sense, getting sieved for free since we'll want the rest of the file sieved more. The heuristic I've found is that sieving until the sec/factor is at least double the LLR time for the smallest candidate is at the lower end of the right time to break off a piece.
However, you won't be LLRing it yourself- BOINC will. Just like the NFS@home matrix-solvers have NFS@ home make extra relations to make their matrix task shorter, there's nothing wrong with sieving less for BOINC than you might for yourself. For Riesel-3, I've been sending pieces of the sieve files to rebirther when sec/factor in sieve is the same as sec/test in LLR for a typical candidate in the piece.
30e12 would be a good spot to aim for for BOINC work, 50-75e12 for your own work.
2015-12-13, 11:37 #13
LaurV
Romulan Interpreter
Jun 2011
Thailand
24×13×47 Posts
Quote:
Originally Posted by VBCurtis If you test the 300k-1M sieve speed vs 200k-1M, you'll find that there's very little speed gain, meaning the 200-300 piece is, in a sense, getting sieved for free since we'll want the rest of the file sieved more.
Thanks for the reply, I found that, or something similar. I got the idea that I need to sieve to 3e13, that is what I need to know. In a sense, as you said, if some prime divides some s(n), the n is "random", so its chance to be in 300k-1M is 7/8 from the chance that n is in 200k-1M, and this is what the factor list shows: about one in 8 is in 200k-300k, and i get one every ~125 seconds, i.e. about 1000 seconds per factor.
2015-12-13, 21:50 #14
Puzzle-Peter
Jun 2009
68610 Posts
Quote:
Originally Posted by rebirther Hi Fellowers, if someone is able to sieve some S/R Bases from 25-100k this would be great. Also 100-200k and 200-400k are welcome. Iam out of running these ranges soon in BOINC and I dont want to loose time. Thx in advance!
Do you also need some XXL tests? I could give you a ready-to-go sieve file for S/R22 n=1.5M - 2M.
2015-12-14, 05:57 #15
rebirther
Sep 2011
Germany
2×1,489 Posts
Quote:
Originally Posted by Puzzle-Peter Do you also need some XXL tests? I could give you a ready-to-go sieve file for S/R22 n=1.5M - 2M.
No, the main target is 25-100k
2015-12-15, 19:57 #16
wombatman
I moo ablest echo power!
May 2013
23×223 Posts
R347
Here's the sieved file (P=20e12) for R347 from n=200k-400k. There are 1679 candidates remaining.
Edit: I can go ahead and run base 373 as well. Is there an easy to run both the S373 and R373 at the same time? Or do I just run them concurrently with two instances of the siever?
Nevermind, got it sorted.
Attached Files
r347_20e12.zip (4.2 KB, 238 views)
Last fiddled with by wombatman on 2015-12-15 at 20:21
2015-12-16, 01:40 #17 LaurV Romulan Interpreter Jun 2011 Thailand 100110001100002 Posts S428 status report: reached 12T, continuing to 30T as per Curtis' advice. In the same time I let cllr running for S428(8), it reached n=120k (this is a DC, but I don't think I will go to the full n=200k, as it is already getting very slow).
2015-12-21, 23:28 #18 wombatman I moo ablest echo power! May 2013 23×223 Posts S/R373 and R610 R/S373 (k=18 and k=108, respectively) is on its way to P=20e12 and should finish Christmas Day. R610 is heading to P=15e12, finishing on the 23rd.
2015-12-22, 00:41 #19 gd_barnes May 2007 Kansas; USA 23·5·263 Posts Later tonight, I'll include a list of what is being worked on and who has what reserved for BOINC sieving in the first posting here. I'll also put a link to all completed files on the pages. Ian just sent me a large batch of bases sieved for n=25K-100K. That should keep BOINC busy for a while. :-)
2015-12-22, 04:00 #20 wombatman I moo ablest echo power! May 2013 23·223 Posts Sweet! Thanks!
2015-12-23, 16:57 #21
wombatman
I moo ablest echo power!
May 2013
23×223 Posts
R610
Here's R610 for n=200k-400k, taken to P=15e12. 2284 candidates remaining.
Attached Files
r610_15e12.zip (6.1 KB, 313 views)
Last fiddled with by wombatman on 2015-12-23 at 16:57
2015-12-25, 06:11 #22 gd_barnes May 2007 Kansas; USA 101001000110002 Posts I received a large batch of sieve files from Ian for n=25K-100K and 100K=200K. There is now a link to all of them on the pages. They are as follows: Code: R91 100K-200K R166 25K-100K R291 25K-100K R313 25K-100K R403 25K-100K R418 25K-100K R421 25K-100K R490 25K-100K R510 100K-200K R553 25K-100K R565 25K-100K R602 100K-200K R694 100K-200K R696 100K-200K R783 100K-200K R803 100K-200K R823 25K-100K S91 100K-200K S166 25K-100K S243 25K-100K S421 25K-100K S490 25K-100K S591 25K-100K S820 25K-100K S821 100K-200K S823 25K-100K S832 100K-200K S931 25K-100K
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2021-10-18 07:29:25
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https://mathoverflow.net/questions/116051/localizations-of-non-nilpotent-spaces
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# Localizations of non-nilpotent spaces
For simplicity let's talk about $p$-localizations of spaces for a fixed prime $p$. Every space $X$ has a well-defined $p$-localization which can be constructed by the small object argument and which becomes a fibrant replacement in the $p$-local model structure on the category of spaces. It is well-known that nilpotent spaces have nice enough Postnikov towers and we can localize such spaces by taking the Postnikov tower, localizing step by step and putting it back together by taking the limit of the resulting tower of fibrations. My question is:
Is there an example of a non-nilpotent space $X$ whose $p$-localization we can explicitly describe?
I leave the meaning of "explicitly" ambiguous. I would be interested in any construction not using the small object argument.
Here's my stab at a possible example. For a group $G$ we define its lower central series by setting $G_0 = G$ and $G_{n + 1} = [G_n, G]$ and we can also continue transfinitely by setting $G_\beta = \bigcap_{\alpha < \beta} G_\alpha$ for limit ordinals $\beta$. The group $G$ is nilpotent if this construction terminates at the trivial subgroup at a finite stage. It is called hypocentral if it terminates at the trivial subgroup at some not necessarily finite stage. According to Wikipedia it is a result of Malcev that there are hypocentral groups with arbitrarily long lower central series.
If we start with a hypocentral group $G$ and convert its lower central series into a (transfinite) tower of fibrations (whose limit is a $K(G, 1)$), $p$-localize it step by step and take the limit of the resulting tower of fibrations, do we obtain the $p$-localization of $K(G, 1)$?
• Here is one example, albeit too simple to be very interesting: The projective plane RP^2 is not nilpotent, and is the cofibre of the degree-2 map S^1 -> S^1. If you invert 2, then this becomes an equivalence, so its cofibre is contractible. Since localization is a left Quillen functor (or left adjoint on the infinity-level) it preserves homotopy colimits, so the localization of RP^2 at any odd prime is contractible. – Rune Haugseng Nov 1 '17 at 12:21
As you know extending $P$-localization functors to non-nilpotent spaces is a very delicate matter. You can do it by homotopical localization techniques. if you do it in this way you get functorial localization which work for non-nilpotent spaces "but" it is the aim of your question it is not "very explicit".
Now let me be very naive and let us go back to D. Sullivan's MIT notes "Geometric topology, localization, periodicity and Galois symmetry" chapter 2 (these notes are really beautiful to read). D. Sullivan first explains how to do $P$-localization when we have a CW-complex, of course he also explains that this CW-procedure gives a "nice localization" when the CW-complex is 1-connected (in that case it is our modern $P$-localization). He proceeds on the skeleton of the CW-complex, localizing cell by cell. Thus you can apply Sullivan's telescope construction to a circle and then to a wedge of circles and any CW-complex you want. The starting point being the localization of spheres through a telescope of maps $S^k\stackrel{\times p}{\rightarrow }S^k$ (you invert degree $p$-maps).
The point is that you can argue that this naive construction on non-nilpotent CW-complexes is not what you expected as a $P$-localization. D. Sullivan also explains how to do it when we have a Postnikov tower.
Let me add that there is another interesting way to $P$-localize which is due to C. Casacuberta and G. Peschke: "Localizing with respect to self-maps of the circle" Trans AMS 339 (1993) 117 – 140.
This $P$-localization is as "explicit" as a homological Bousfield's localization but the homotopy related to it is very interesting. In that setting, a space is $P$-local if and only if the $p$-power map on the loop space $\Omega X$ is a self homotopy equivalence if $p\notin P$.
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2019-12-12 15:41:29
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https://math.stackexchange.com/questions/4046747/integral-of-e-at-erft-and-e-at-erft2
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# Integral of $e^{-at}$ Erf[t] and $e^{-at} Erf[t]^2$
Can anyone help me with any suggestions for the following triple integral?
$$\frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^t dt_1 e^{-(\gamma_1+\gamma_2)t_1} \int_{-\infty}^{t_1} dt_2 e^{\frac{\gamma_1+\gamma_2}{2}t_2} e^{-\frac{t_2^2}{4\sigma^2}} \int_{-\infty}^{t_2} dt_3 e^{\frac{\gamma_1+\gamma_2}{2}t_3} e^{-\frac{t_3^2}{4\sigma^2}}$$ It can be put into the equivalent form, $$\frac{1}{2\sqrt{2\pi \sigma^2}}\int_{-\infty}^t dt_1 \left[ \left[ e^{-\frac{\gamma_1+\gamma_2}{2}t_1} \Theta(t_1) \right] * e^{-\frac{t_1^2}{4\sigma^2}} \right]^2$$ where $$\Theta(t_1)$$ is the Heaviside unit step function and * denotes convolution.
This second integrand can be written in terms of he error function, Erf[],
$$e^{\frac{(\gamma_1+\gamma_2)^2 \sigma^2}{2}} \frac{1}{2\sqrt{2\pi \sigma^2}}\int_{-\infty}^t dt_1 e^{-(\gamma_1+\gamma_2)t_1} \left[ \frac{1}{2} + \frac{1}{2}Erf\left[\frac{t_1-(\gamma_1+\gamma_2) \sigma_1^2}{2\sigma} \right] \right]^2$$ The definition of Erf[] being $$Erf[\frac{t-t_0}{\sigma}] = \int_0^{t-t_0} dt_1 \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{t_1^2}{2\sigma^2}}$$
Mathematica fails with the computation, even numerically, in the last part.
As asked in title $$I=\int e^{-a t}\, \text{erf}(t)\,dt$$ is simple (just one integration by parts and completing the square) $$I=\frac{e^{\frac{a^2}{4}}\, \text{erf}\left(t+\frac{a}{2}\right)-e^{-a t}\, \text{erf}(t)}{a}$$
But the second one $$J=\int e^{-a t}\, \big[\text{erf}(t)\big]^2\,dt$$ seems to be much more problematic.
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2021-12-07 03:31:13
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http://www.techhelpmanual.com/292-shell___config_sys_command_.html
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# SHELL= (CONFIG.SYS Command)
``` Compatibility: 2.0+
Purpose: Replaces the normal command processor (COMMAND.COM) with a
different program, OR selects non-default settings for
COMMAND.COM. Note: This command is unrelated to the Dosshell
command.
Uses: ■ Identify the location of COMMAND.COM.
■ Create a DOS environment that is larger than the default.
■ Run a batch file other than AUTOEXEC.BAT at startup.
Syntax: SHELL=[d:][path]filename.ext [parms]
█▌Parameters▐█
d:\path\filename.ext is the drive, path, filename and extension of the
program to be used as the command processor.
progParms specifies command-line parameters for the command processor.
█▌TECH Notes▐█
■ The DOS 6.0 COMMAND.COM option /K= sets the name of the AUTOEXEC.BAT
file. COMMAND.COM /K=mybat.bat is a handy way to set up a DOS session
from Windows.
■ To increase the size of the DOS environment, use COMMAND.COM /E=nnn.
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2018-12-17 15:03:58
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http://map.grauw.nl/articles/yjk/
|
# The YJK screen modes
An exploration of the YJK screen modes of the MSX2+.
## Introduction
The Yamaha V9958 VDP used in the MSX2+ and MSX turbo R adds two high colour display modes which use a different colour encoding than traditional RGB, called YJK and YJK+YAE. These modes expand the colour count to 19268 and 12499 + 16 respectively, at the cost of colour attribute clash between groups of 4×1 pixels.
The YJK mode is MSX-BASIC’s SCREEN 12. It can show up to 19268 colours, and is particularly suited for photographic images because these are generally fairly unaffected by colour attribute clash. See art gallery.
The YJK+YAE mode is MSX-BASIC’s SCREEN 10 and 11. It can show up to 12499 YJK colours, and 16 palette colours (out of 512). Since the palette colours are set per-pixel they can be used to hide colour bleeding, while the YJK colours can be used to make the picture more colourful. Due to this YJK+YAE is more versatile in situations when you also want to display a user interface, like in a game. See art gallery.
An additional benefit of these new screen modes have compared to SCREEN 8 is that their sprites make use of the palette, rather than a limited set of predefined colours.
## Selecting the YJK modes
On the VDP the YJK mode is selected by following the procedure to select SCREEN 8 (G7), and then setting the YJK bit of register 25. To select the YJK+YAE mode, additionally set the YAE bit.
Use the CHGMOD BIOS call to switch to SCREEN 8 and then read the r#25 mirror RG25SA, set the YJK and YAE bits in r#25 appropriately and write the register back with WRTVDP.
Since a number of MSX2 computers exist that do have a V9958 but no BIOS support, as well modded MSX2 computers with their VDP replaced, consider setting r#25 directly rather than through the BIOS if the MSX version number is 1 (MSX2) and the VDP ID in s#1 bits 1-5 is 2 (V9958). In this case the other r#25 bits can be assumed 0.
In MSX-BASIC, SCREEN 12 and 11 have no special support for drawing in YJK other than using 8-bit colour operations. However SCREEN 10 is special, it gives a palette-colour view of YJK+YAE and allows you to draw on it like in SCREEN 5. Switching between SCREEN 10 and 11 will not clear the screen.
## The encoding of YJK
YJK breaks down into three components. The Y component specifies brightness (luma) and a little bit of blue. The J and K components contain most of the colour information (chroma), expressed as a difference between Y and red and green, respectively. The concept behind this encoding is that the human eye is more sensitive to brightness than colour, and the colour resolution can therefore be reduced with relatively limited impact on the perceived result.
Use the visualisation widget above to get a feel of the YJK colour space. You can interact with this widget to view the various colour ramps available. The horizontal axis is the J value, the vertical axis is K. The column on the right controls the Y amount. All colours in the Y column can be used in the same four-pixel group. As you increase Y the colour becomes brighter, but notice that it becomes slightly more blue as well. This can be seen well in the grayish hue when both J and K are zero.
In YJK mode the VDP stores the J and K components in 6 bits as a two’s complement signed value (-32 to 31). These are shared between 4 adjacent pixels. The Y component is a 5-bit unsigned value (0 to 31). This one can be specified per-pixel.
In YJK+YAE mode the VDP still stores the J and K components in 6 bits as a two’s complement signed value (-32 to 31). However the Y component can now only specify even values (0, 2, 4 .. 30), as its least significant bit is used for the attribute bit A. If the A bit is set, the top four bits of the Y component specify a palette colour to use instead.
The palette is still specified as 9 bits, with 3 bits per component. However the V9958’s DAC is 15-bit, so there must be some kind of mapping in place. Indeed, the 3-bit RGB palette component values map to 5 bits as follows:
This applies to all screen modes, by the way.
## Conversion between YJK and RGB
The Y, J and K values specify 17 bits of information in total. The V9958 converts these to 15-bit RGB with 5 bits per colour component. The V9958 manual describes the following conversion formulas for YJK. The formulas for YJK+YAE are the same, but the least significant bit of Y is always zero so all its values are even.
Indeed those are fine formulas and perfect mathematical inverses of each other. However they do not take into account that the VDP both rounds and clips the result of the YJK to RGB conversion.
### Rounding
Let’s consider the rounding first. The y, j, k and r, g, b values are all integers, so when the blue value is calculated, it is rounded down to the nearest integer value (“floor”). To compensate for this in the RGB to YJK formula, you need to round up (“ceil”) the value resulting from the y.
Using these formula you can express exactly half of what 15-bit RGB allows; 16384 out of 32768 colours (50%), and 8192 for YJK+YAE (25%). Each representable RGB value maps to one unique YJK value and vice versa. If you want to keep things simple, stop here and stick to these 16384 colours.
### Clipping
As we know though YJK mode can generate 19268 colours. Where did the remaining 2884 colours go? YJK’s 17 bits allow for 131072 different values, but most of these fall outside the range of 15-bit RGB. This is called “out of gamut”. Values that are out of range are clipped to [0, 31].
As a side effect of this clipping, 2884 colours which could not be represented by the previous YJK formulas become available. For example, take the colour (0, 24, 31). Without clipping this could not be expressed, since applying the previous RGB to YJK to RGB formula will yield (0, 24, 32). Due to the clipping of the blue component though, this colour is available to us after all.
If we calculate the number of colours affected by clipping, those which have one component of either 0 or 31, there are 323 - 303 = 5768 of them. Half of them are already covered by unclipped colours, 5768 / 2 = 2884, meaning that all of the colours where any RGB component is 0 or 31 can be represented. Some of these have only one YJK representation, others up to 3463.
You may notice no RGB to YJK formula is specified. This is because although colours whose components all fall within the [1, 30] range have a unique YJK representation which can be determined with the earlier formula, colours with one or more components of either 0 or 31 have multiple solutions.
In order to deal with this, consider that an RGB value is a point in a 3-dimensional (colour) space. For each potentially clipped component (0 or 31) a ray in the clip direction extends from the point, where two rays form a bounded plane, and three rays a volume. This colour space can be transformed to YJK with a matrix, and all YJK values overlapping this point, ray, plane or volume represent the same RGB colour.
More on this in part 2, which will be a deep dive into the topic of image conversion.
## Drawing techniques
The YJK modes can be very nice for pixel art because of the wider range of colours that you an express with 15-bit RGB. Ever wanted that particular pastel colour that the V9938’s 9-bit RGB palette could not express? Or a smoother gradient? The V9958 can do in YJK. Additionally the high colour count allows you to create very colourful artwork.
However on the flipside the colour restrictions of YJK make it a difficult mode to work with. Below you will find some tips on how to work with the constraints of the YJK modes.
For pixel art I recommend to use the YJK+YAE mode. The availability of palette colours greatly increases your flexibility to work around the colour clash restrictions. The YJK mode is more suited for photographic material with smoother gradients and lower contrast.
The simplest technique is to approach the art like you would a SCREEN 5 image. Draw the majority of the image using the 16 palette colours. Consider it a layer on top of the YJK layer. Then “punch holes” through this layer by introducing YJK colours in select places, adhering to the 4×1 pixel group restrictions. The easiest way to do this is by simply using only a single colour per pixel group. This already allows you to introduce many more colour details without needing extra palette colours for them, while avoiding the complexities of YJK.
If you want to use multiple YJK colours within those 4 pixels, you need to pay attention to the colours available in the specific J, K combination for that quad. This requires a bit of planning, you’ll want to start by establishing a few useful colour ramps to reuse them througout the art piece. Choose colours from an Y ramp you like, for example using the tool at the top of this page, and then note them in a little scratch area for easy colour picking. Then whenever you need to draw e.g. some grassy bits, you can pick from the green ramp you’ve noted, while being mindful of the 4×1 grid.
For a more cartoony drawing style in YJK mode, you could draw the artwork in grayscale, and then paint the colour on a separate colour modifier layer with a blurry brush, so that there are no hard colour transitions. This will convert to YJK well.
• Align the vertical edges in your graphics to the 4×1 grid to minimise the effect of colour bleed.
• Use the out of gamut colours of the Y ramps to your advantage, their hue changes more than usual so they give some extra flexibility.
• In situations where you can’t use the colour you want, try to match its brightness so that the colour difference is harder to spot.
• The eye is sensitive to edges, so de-block visible edges of colour bleeding by adding detail or dithering in them.
In the end to which degree you want to apply these techniques depends on how much time you want to invest. Please share your experiences and tips and tricks in the msx.org pixel art thread!
## Comparison to Y′UV
The YJK colour space is similar to the more common Y′UV colour space, with 4×1 chroma subsampling like the Y′UV411 encoding. However they differ in the weights applied to the different colour components. Where Y′UV assigns weights to make the Y component express luma (brightness), YJK has swapped the weights for green and blue, meaning that Y does not only affect luma but also the amount of blue.
This trait can be a bit unfortunate, as in natural scenes the amount of blue decreases under (sun)light, and increases in shade. Y′UV would work better in those situations. YJK is perhaps better suited for metallic, fluorescent and pastel colours.
In an interesting article “Issues on YJK colour model implemented in Yamaha V9958 VDP chip”, Ricardo Cancho Niemietz elaborates why this is not ideal from the perspective of the luminance-chrominance colour model, and suggests that it must be an unintentional mistake of the Yamaha engineers. Assuming though that Yamaha’s engineers were smart people, and wouldn’t make a colour model inspired by YUV but deviate in such a detail without reason, let’s try to find that reason.
In YJK’s 4-byte encoding there are six bits for J and K, but only five for Y. Due to this it can only represent half of the RGB colours, 214 instead of the 215 that 15-bit RGB can produce. Ideally Y would have 6 bits as well to get the full range, but this doesn’t fit in the byte encoding (and certainly not in YJK+YAE). So which RGB colour component’s resolution do we reduce by one bit?
The most logical answer here is blue, just like in SCREEN 8. Because our eyes are the least perceptive of blue, we won’t notice it as much. And indeed in YJK blue is effectively only 4-bit, in YJK+YAE effectively 3-bit. Had Yamaha gone for YUV, green would have gotten the least resolution, while it is the most visible colour, causing more banding in it. From this perspective, YJK spends its bits more effectively on the colours our eye perceive best.
A final note on this; the Yamaha V9990 VDP has both an YJK and an YUV mode, so on that VDP you can use the mode of your choosing. The YUV mode works identical to YJK on the V9958 but with the weights for the G and B components swapped. The weights do not perfectly match the real Y’UV but it’s pretty close.
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2023-01-28 14:10:27
|
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http://www.perlmonks.org/index.pl?node_id=80677
|
laziness, impatience, and hubris PerlMonks
### Re: Re: Re: Re: Re: Fastest way to compare multiple variables?
by mikfire (Deacon)
on May 15, 2001 at 23:10 UTC ( #80677=note: print w/replies, xml ) Need Help??
Okay, I cannot resist. This is the product of not enough sleep and I warn all who continue. Given we have used merlyn's suggestion for a hash of arrays ( I was actually thinking of an array of ararys, but why not a hash? ), what if we did something like this
sub compare {
my %data = @_;
my @lengths = sort { $a <=>$b } map { scalar@{$data{$_}} } keys %da
+ta;
return $lengths[0] ==$lengths[-1];
}
[download]
Basically, if we sort the lengths and the last element is equal to the first element, then everything else inbetween must be equal.
This has likely no value - it really isn't clearer nor does it likely save any cycles. I just thought it was fun and have not had enough sleep.
UPDATE: did I mention not enough sleep? Fixed a typo
mikfire
Create A New User
Node Status?
node history
Node Type: note [id://80677]
help
Chatterbox?
[thezip]: Is there an analogy for '&' (ie. run commandline process in background) for Windows commandline? [Corion]: thezip: start "some title" path\to\that\ application, but that will open another console window [Corion]: thezip: If you want to confuse your users, use system(1, "that\\command" );, which will make Perl launch it in the background [Corion]: That will keep the console window open even though the user can't type into it anymore
How do I use this? | Other CB clients
Other Users?
Others exploiting the Monastery: (14)
As of 2017-03-27 18:46 GMT
Sections?
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Leftovers?
Voting Booth?
Should Pluto Get Its Planethood Back?
Results (321 votes). Check out past polls.
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2017-03-27 18:46:06
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https://collegephysicsanswers.com/openstax-solutions/write-complete-decay-equation-given-nuclide-complete-aztextrmxn-notation-refer-0
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Question
Write the complete decay equation for the given nuclide in the complete $^A_Z\textrm{X}_N$ notation. Refer to the periodic table for values of Z:
$\beta^+$ decay of $^{50}\textrm{Mn}$
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2019-10-16 04:16:38
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{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8996492624282837, "perplexity": 1066.523036943632}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986664662.15/warc/CC-MAIN-20191016041344-20191016064844-00078.warc.gz"}
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http://math.stackexchange.com/questions/114278/a-group-of-order-66-has-an-element-of-order-33
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# A group of order 66 has an element of order 33
If I wanted to show that a group of order 66 has an element of order 33, could I just say that it has an element of order 3 (by Cauchy's theorem, since 3 is a prime and 3|66), and similarly that there must be an element of order 11, and then multiply these to get an element of order 33? I'm pretty sure this is wrong, but if someone could help me out I would appreciate it. Thanks.
-
This generally isn't true unless the two elements commute. – Qiaochu Yuan Feb 28 '12 at 3:48
I figured this would be the case. Is there a way to repair my "proof"? – johnq Feb 28 '12 at 3:50
Show that they commute! – Mariano Suárez-Alvarez Feb 28 '12 at 3:52
It's not clear that they do necessarily commute. – johnq Feb 28 '12 at 4:01
I did not say it was clear, I said prove it :) – Mariano Suárez-Alvarez Feb 28 '12 at 4:02
Step 1: Show that any finite group order $4k+2$ has an index two subgroup. (Hint.)
Step 2: Show that any group of order $33$ is cyclic.
-
Alternatively:
• show that the Sylow $11$-subgroup is normal.
• show that a cyclic group of order $11$ has no automorphisms of order $3$.
• pick an element of order $3$ in your group and conclude that it commutes with any element of order $11$.
-
To compliment the other answers, I will address why commutativity is necessary. Let $G$ be your group $o(x)$ denote the order of $x\in G$. You are making use of the statement that $o(a)o(b)=o(ab)$, but this is not necessarily true. It holds when $o(a),o(b)$ are coprime and $ab=ba$ (an interesting consequence of this is that the partial sums $s_n$ of the harmonic series are never integers for $n>1$, which follows from Bertrand's postulate and $\mathbb Q/\mathbb Z$ being abelian), but there are nonabelian groups of order $66$, such as $S_6\oplus \mathbb Z_{11}$. If $ab=ba$ and $o(a),o(b)$ are coprime, then $(ab)^{o(a)o(b)}=a^{o(a)}b^{o(b)}=e\cdot e=e$, so $o(ab)\leq o(a)o(b)$. If $(ab)^n=e$ then $a^nb^n=(ab)^n=e$ so $(a^n)^{-1}=b^n$, hence $o(a^n)=o(b^n)$, and if $n<o(a)o(b)$ then since $o(a),o(b)$ coprime we have that one of $a^n,b^n\neq e$. But $o(b^n)=o(a^n)|o(a)$ since $(a^{n})^{o(a)}=(a^{o(a)})^n=e^n=e$, and similarly $o(a^n)=o(b^n)|o(b)$, so $o(a^n)=o(b^n)=1$. Thus $a^n=b^n=e$, so $n\geq o(a)o(b)$ so $o(ab)=o(a)o(b)$.
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2014-03-11 00:34:36
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https://runescape.wiki/w/Duradel
|
Duradel is the fourth highest level Slayer Master and the father of Kuradal. Players must have at least level 50 in the Slayer skill, 100 combat and have completed the Shilo Village quest before they may access him and receive Slayer assignments from him. Duradel is found on the 1st floor[UK]2nd floor[US] north of Fernahei's Fishing Hut in Shilo Village. Alternatively, if Lunar Diplomacy is completed, the spell NPC Contact can be used to speak and get assignments from him.
He perishes during the events of While Guthix Sleeps and is replaced by Lapalok as Slayer Master.
## History
### Early life
Duradel excelled in Slayer and combat, and aspired to become the greatest Slayer master. He took a potentially questionable path, the details which he does not reveal to prevent others to be tempted, to reach his goal. Duradel eventually successfully became the most powerful Slayer master, and remained so for a time, but became trapped in the role and was not allowed to leave.[1][2]
Duradel trained Vannaka, his talented student, who would eventually rise to the rank of Slayer master in his own right. At some point, Vannaka challenged Duradel to be the strongest Slayer master, only to be defeated by him with a rare and powerful weapon. Vannaka accepted defeat, and Duradel remained the most powerful Slayer master then.[3]
Duradel married and had a daughter, Kuradal. Kuradal was very close to her father, and from a young age, was trained by him in combat and Slayer skills. Kuradal became very proficient in Slayer and combat, and also quite competitive, aspiring to surpass him. Although Kuradal assigns harder slayer creatures than her father, she still thought he might be more powerful and isn't sure who was stronger.[4] Kuradal later trained another master, Morvran, an elf of the Iorwerth Clan, who offers even greater Slayer challenges than her own, following the restoration of Prifddinas.[5]
At some point, Duradel separated from his wife. Kuradal had been making preparations for bringing her mother and father back together, prior to Duradel's death.[6]
### Fighting against Lucien
The following takes place during While Guthix Sleeps.
Duradel was one of eight warriors recruited by an adventurer to hunt Lucien. After his recruitment, Lapalok replaces him in Shilo Village as the third highest-level Slayer master. His replacement was intended to be temporary at first, as Duradel would be unable to perform his usual Slayer master duties while at war. Lapalok, as the Disciple of the Devourer, seemed quite eager to send Duradel and the adventurer away to war, where he expected both would meet destruction.
Upon questioning him in the White Knights' Castle as he prepares for his fight, he reveals he was being kept in Shilo Village by an unknown higher authority, and required special permission from them–he seemed to communicate telepathically with them–to leave for his part in the battle against Lucien.[1] He reveals that, apparently, he took several shortcuts and 'trod on very important toes', as he puts it, to become the greatest Slayer master–which he was at the time, before being surpassed by his daughter, Kuradal. Unfortunately, he would not reveal any more information to the adventurer, and stated that he will take this secret to the grave to avoid tempting others including the adventurer.[2] He does just that.
The group attempts to save the adventurer from Lucien in the Wilderness. They engaged in battle with Lucien and his undead army. It is revealed during the battle that Duradal has a combat level of 197. Duradel is one of the six who perish in the ensuing battle, all six of whom were fighting Lucien directly. Lapalok then replaces him permanently as Slayer master.
After this quest, a statue of Duradel is fittingly placed next to a statue of Turael at the entrance to the Slayer Tower near Canifis, and two more statues can be found in Falador along with the other five heroes: around the lake in Falador Park and lining the bridge to the White Knights' Castle.
Duradel's soul makes a small appearance in the Underworld, right outside Icthlarin's castle where he wistfully ponders his time as a slayer master and his wife.
## Slayer points
• 15 points per normal task (7 if Smoking Kills hasn't been completed)
• 75 for every 10th (37 otherwise)
• 225 for every 50th (115 otherwise)
This means that if using Duradel over the long term, an average of 24 points will be granted per task.
## Assignments
1. ^ Requires you to speak to the dying knight outside of God Wars Dungeon to unlock them as a task.
2. ^ Jagex. Mod Daze's Twitter account. 7 August 2018. (Archived from the original on 20 December 2019.) Mod Daze: "Talk to the guy on the surface and it should unlock then"
3. ^ Players need to own a fire cape or TokHaar-Kal (equipped, in player-owned house, in bank, or in inventory) or have purchased the ability to get an ice strykewyrm Slayer task for 2,000 Slayer points. Players must also have completed The Tale of the Muspah. Dropping all fire capes before obtaining a new task and picking them up afterwards can avoid this assignment.
### Slayer Challenge
After a player has performed several Slayer tasks for Duradel, he will occasionally offer a special task in place of a regular Slayer task. The player can decline to take the special task without penalty. If the player accepts the special task, completing the task:
• Earns extra Slayer experience
• Earns Slayer points
• Counts as a task for the player's consecutive task count, which is used for Slayer points.
Duradel's special task is for the player to slay monsters in the Slayer Tower in a single trip without leaving the tower. The numbers of monsters that must be slain are:
• 17 Crawling Hands on the ground floor[UK]1st floor[US]
• 12 Banshees on the ground floor[UK]1st floor[US] (requires protection from the banshees' screams)
• 9 Infernal Magi on the 1st floor[UK]2nd floor[US]
• 7 Bloodveld on the 1st floor[UK]2nd floor[US]
• 7 Aberrant spectres on the 1st floor[UK]2nd floor[US] (requires protection from the spectres' odour)
• 5 Gargoyles on the 2nd floor[UK]3rd floor[US] (requires a Rock hammer)
• 6 Nechryaels on the 2nd floor[UK]3rd floor[US]
• 6 Abyssal demons on the 2nd floor[UK]3rd floor[US]
As the player kills the required number of each set of monster, a notice appears in the chatbox that the player has killed enough of the monster for the challenge.
Note that level 85 Slayer is needed for this task to kill every monster in the Slayer Tower. All monsters will use either magic or melee attacks so either Ganodermic armour or the use of prayer can easily destroy most of the monsters. A Full slayer helmet and the ability to destroy specific monsters would be highly recommended.
The reward for completing this task is a bonus of 20,000 Slayer experience and 35 bonus Slayer points.
Note: Leaving the Slayer Tower and failing to return within the given time frame will cancel the special assignment but will not reset the player's slayer task count.
A dynamic calculator for task weight can be found here.
The percentage chance of getting assigned a given Slayer task can be calculated using the formula
${\displaystyle {\frac {w}{S}}\times 100\%}$
where ${\displaystyle w}$ is the task's weight and ${\displaystyle S}$ is the sum of all weights for the particular Slayer Master. Note that the weights of all blocked tasks, as well as tasks toggled off have to be subtracted from the sum ${\displaystyle S}$. For Duradel, assuming all possible assignments are available, ${\displaystyle S}$ = 381.
Preferring a task makes it approximately twice as likely to be assigned, for the first assignment choice, but does not affect the weighting of the second assignment choice provided by a Slayer VIP ticket.[7]
## Update history
The update history project is a work-in-progress – not all updates to this topic may be covered below. See here for how to help out!
• patch 10 August 2015 (Update):
• A skin tone issue with Duradel and Lapalok has been fixed.
• patch 17 February 2014 (Update):
• patch 24 November 2010 (Update):
• The Lapalok/Duradel slayer challenge will begin upon entering the Slayer Tower.
• patch 18 October 2006 (Update):
• Higher-level slayers may be pleased to hear that Duradel, the Slayer Master of Shilo Village, has finally restored Nechryael to his list of assignments.
• patch 20 September 2006 (Update):
• Duradel, the Slayer Master in Shilo Village, has been persuaded to rethink the assignments he sets. He's dropped the weaker monsters from his list, so experienced adventurers may find him much more likely to set them a challenge worthy of their combat skills.
• update 26 January 2005 (Update):
• Today we have introduced a new skill for members called the Slayer skill! To start the Slayer skill, you should find one of the Slayer Masters who are located in various places around RuneScape. There are five Slayer Masters in total, each catering to different strengths. They can be found in Burthorpe, Morytania, Edgeville Dungeon, Zanaris, and Shilo Village.
## Trivia
A statue in memory of Duradel
• When asked about Vannaka in a Postbag from the Hedge, it is revealed that Vannaka was a student of Duradel's, and once challenged him to be the strongest Slayer Master only to be beaten by his master with a sort of rare and powerful weapon.[3]
• On the day of the release of While Guthix Sleeps, some Slayer masters, including Duradel, had a glitch in their names in the chatbox. If a player was assigned a task and asked "Got any tips for me?", instead of Duradel or Lapalok the game showed slayer_master_5_multi. This glitch could not be seen if using an enchanted gem, slayer ring or the NPC Contact spell. It could only be seen if talking personally to Duradel or Lapalok.
• He was the toughest Slayer master until the release of Kuradal on 8 December 2009. Prior to Kuradal's release, he wore and sold the Slayer cape, but he now wears a cape of legends and Kuradal sells the Slayer cape.
## References
1. ^ a b Duradel, "While Guthix Sleeps", RuneScape. "Good to be out of that hut I've inhabited for the last few years. It seems that saving the world was enough reason for me to be allowed out."
2. ^ a b Duradel, "While Guthix Sleeps", RuneScape. "Let's just say that you don't arrive as the greatest slayer master overnight without a lot of blood and slaughter. I took some shortcuts and trod on some very important toes... [the method] is a secret that will go to my grave. I cannot take the chance that you would be tempted as I was."
3. ^ a b Jagex. Postbag 20 - "Transcript:Tales from earlier ages", Letter 8, by Duradel. RuneScape Postbags from the Hedge.
4. ^ Kuradal, "Festival of the Dead", RuneScape. "Whenever I think of my father, I return to my childhood, when we would slay dangerous beasts together. We were both far too competitive for that to last long. I would go on my own at night and slay tougher and tougher creatures, so that I could surpass him. Now we will never know who was the better slayer."
5. ^ Morvran, "Plague's End", RuneScape. "In my travels, I made the acquaintance of a lovely young lady by the name of Kuradal. She taught me much about the wider world... I found myself fascinated by her and what she does, and so I decided to turn my talents and knowledge towards slaying."
6. ^ Kuradal, "Festival of the Dead", RuneScape. "I wished to bring mother and father back together. I had completed many preparations. I should have done it sooner."
7. ^ Jagex. Mod Rowley's Twitter account. 26 September 2014. (Archived from the original on 28 September 2014.) Mod Rowley: "Preferred Slayer tasks = a silent, second random roll. If either task is on your prefer list, that's what you get (bias towards first roll)... Once normal and prefer roll is decided, [if] you have a VIP ticket, you get the choice between that roll and a third roll."
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2020-09-29 17:51:47
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https://gamedev.stackexchange.com/questions/32692/platformer-movement-and-collisions
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Platformer movement and collisions
I'm working on a platformer but I'm stuck at the player movement and collisions.
Right now I'm not including the grid at all in the movement, but it's time I do it. I've read this article on how to do it. But there's no real code examples just theory.
I like the second example and that's the one I'm going for. I want the movement to be smooth, and the player should be able to run in between squares from my map array. This means I have to get tile information from the array, and according to that allow/forbid my player to move in that direction. And that's where I'm stuck.
My array is a bunch of 1's and 0's right now. I know how to calculate a hitbox around the player, but how do I compare that to the array in an efficiant way?
UPDATE: I've been playing around with this for a while now, and the only thing I came up with is using player X and Y which is in pixels, dividing by tile size, and getting a rough estimate on the map. Then I make it an AABB by adding its width and height to get the corners.
This showed to be very unreliable though, and I would really need some help on how to structure this.
Well you have to obviously loop through tiles to see what ones are interesting the player and deal with it accordingly.
If your map is not very large it's trivial to loop through all tiles to find the collision. something like tiles.forEach(function(tile){}). Once your game and maps start to get larger this of course can be very intensive.
One solution I can think of is to only check the tiles that in the view port. Then all it becomes is a matter of finding what section your character is in and looping though only the tiles in view.
Another way would be to split your tile map into sections, so section 1 might be (0,0) to (20,20). You then determine what section the charter is and loop through only those tiles. That way you can loop through an even smaller number of tiles
• It crossed my mind, but not as obvious as you try to say. In a platformer I could just go "if direction east check X+1, X+2" or whatever and see what is on that position of the array. Anyway, the collision detection isn't really my problem, the problem is more calculating the player position (which is in pixels) to the level array, if you see the difference. I don't know how to piece the two together in an efficient way. – justanotherhobbyist Jul 21 '12 at 8:05
• @hustlerinc I think you need to have him have a position relative to the view port and relative to the level. – Clay Ellis Murray Jul 21 '12 at 8:11
• Yeah that is kind of the whole point of my question? A good way to structure such a function. – justanotherhobbyist Jul 21 '12 at 21:49
I think you are on the right track. It sounds like you have two subquestions:
• How do I track the position of the player in tilespace and in viewspace?
• How do I efficiently track collisions in tilespace?
In a platformer that I wrote, my position objects had tileX and tileY fields and fineX and fineY fields to help with this exact problem. fineX and fineY were bounded from [0, TILE_SIZE). My physics code took these positions into account when determining what happened to objects as they moved around: first use tileX and tileY to get info from the game map, then use that in combination with fineX and fineY for drawing. I'd love to find out a better way to do this, though.
You're almost there with the collisions, too. The only optimization I might add is to "chunk up" your collision grid a bit. Your "normal" tile is TILE_SIZE x TILE_SIZE, right? So, your collision grid can track (TILE_SIZE*2) x (TILE_SIZE*2) chunks at a time since moust objects, even when crossing tile boundaries, will fit into that space? Maybe even TILE_SIZE*4 to allow for growth or quick moving objects. That way, you decrease the number of lookups to only when your object is crossing chunk boundaries.
Then again, there's a big YAGNI here. When you talk about efficiency, are you actually seeing slowdowns in your game? Modern browsers are pretty good about optimizing for you, especially around array accesses and nested function calls. Don't write too much code around this stuff, complexifying (complecting? anyway...) your object model, until you've profiled a bit or actually seen some slowdowns.
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2019-10-18 23:21:33
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http://www.polyfolds.org/index.php?title=Moduli_spaces_of_pseudoholomorphic_polygons
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# Moduli spaces of pseudoholomorphic polygons
To construct the moduli spaces from which the composition maps are defined we fix an auxiliary almost complex structure $J:TM\to TM$ which is compatible with the symplectic structure in the sense that $\omega (\cdot ,J\cdot )$ defines a metric on $M$. (Unless otherwise specified, we will use this metric in all following constructions.)
Then given Lagrangians $L_{0},\ldots ,L_{d}\subset M$ and generators $x_{0}\in {\text{Crit}}(L_{0},L_{d}),$ $x_{1}\in {\text{Crit}}(L_{0},L_{1}),\ldots ,$ $x_{d}\in {\text{Crit}}(L_{{d-1}},L_{d})$ of their morphism spaces, we need to specify the Gromov-compactified moduli space $\overline {\mathcal {M}}(x_{0};x_{1},\ldots ,x_{d})$. (Here and throughout, we will call a moduli space Gromov-compact if its subsets of bounded symplectic area are compact in the Gromov topology. However, this page will only construct the Gromov-compactified moduli spaces as sets; their topology will be constructed from the Gromov topology.) We will do this by combining two special cases which we discuss first.
## Pseudoholomorphic polygons for pairwise transverse Lagrangians
If each consecutive pair of Lagrangians is transverse, $L_{0}\pitchfork L_{1},L_{1}\pitchfork L_{2},\ldots ,L_{{d-1}}\pitchfork L_{d},L_{d}\pitchfork L_{0}$, then our construction is based on pseudoholomorphic polygons
$u:\Sigma \to M,\qquad u((\partial \Sigma )_{i})\subset L_{i},\qquad \overline \partial _{J}u=0,$
where $\Sigma =\Sigma _{{\underline {z}}}:=D\setminus \{z_{0},\ldots ,z_{d}\}$ is a disk with $d+1$ boundary punctures in counter-clockwise order $z_{0},\ldots ,z_{d}\subset \partial D$, and $(\partial \Sigma )_{i}$ denotes the boundary component between $z_{i},z_{{i+1}}$ (resp. between $z_{{d}},z_{0}$ for i=d). More precisely, we construct the (uncompactified) moduli spaces of pseudoholomorphic polygons for any tuple $x_{i}\in L_{i}\cap L_{{i+1}}$ for $i=0,\ldots ,d$ as in [Seidel book]:
${\mathcal {M}}(x_{0};x_{1},\ldots ,x_{d}):={\bigl \{}(\underline {z},u)\,{\big |}\,{\text{1. - 3.}}{\bigr \}}/\sim$
where
1. $\underline {z}=(z_{0},z_{1},\ldots ,z_{d})\subset \partial D$ is a tuple of pairwise disjoint marked points on the boundary of a disk, in counter-clockwise order.
2. $u:\Sigma _{{\underline {z}}}\to M$ is a smooth map satisfying
• the Cauchy-Riemann equation $\overline \partial _{J}u=0$,
• Lagrangian boundary conditions $u((\partial \Sigma )_{i})\subset L_{i}$,
• the finite energy condition $\textstyle \int _{{\Sigma }}u^{*}\omega <\infty$,
• the limit conditions $\lim _{{z\to z_{i}}}u(z)=x_{i}$ for $i=0,1,\ldots ,d$.
3. The pseudoholomorphic polygon $(\underline {z},u)$ is stable in the sense that the map $u:\Sigma _{{\underline {z}}}\to M$ is nonconstant if the number of marked points is $d+1<3$.
Here two pseudoholomorphic polygons are equivalent $(\underline {z},u)\sim (\underline {z}',u')$ if there is a biholomorphism $\psi :D\to D$ that preserves the marked points $\psi (z_{i})=z'_{i}$, and relates the pseudoholomorphic polygons by reparametrization, $u=u'\circ \psi$.
The case $d=0$ is not considered in this part of the moduli space setup since $L_{0},L_{d}=L_{0}$ are never transverse. However, it might appear in the construction of homotopy units?
The domains of the pseudoholomorphic polygons are strips for $d=1$ and represent elements in a Deligne-Mumford space for $d\geq 2$ as follows:
For $d=1$, the twice punctured disks are all biholomorphic to the strip $\Sigma _{{\{z_{0},z_{1}\}}}\simeq \mathbb{R} \times [0,1]$, so that we could equivalently set up the moduli spaces ${\mathcal {M}}(x_{0};x_{1})$ by fixing the domain $\Sigma _{{d=1}}:=\mathbb{R} \times [0,1]$ and defining the equivalence relation $\sim$ only in terms of the shift action $u(s,t)\mapsto u(\tau +s,t)$ of $\tau \in \mathbb{R}$. This is the only case in which the stability condition is nontrivial: It requires the maps $u:\mathbb{R} \times [0,1]\to M$ to be nonconstant.
For $d\geq 2$, the moduli space of domains
${\mathcal {M}}_{{d+1}}:={\frac {{\bigl \{}\Sigma _{{\underline {z}}}\,{\big |}\,\underline {z}=\{z_{0},\ldots ,z_{d}\}\in \partial D\;{\text{pairwise disjoint}}{\bigr \}}}{\Sigma _{{\underline {z}}}\sim \Sigma _{{\underline {z}'}}\;{\text{iff}}\;\exists \psi :\Sigma _{{\underline {z}}}\to \Sigma _{{\underline {z}'}},\;\psi ^{*}i=i}}$
can be compactified to form the Deligne-Mumford space $\overline {\mathcal {M}}_{{d+1}}$, whose boundary and corner strata can be represented by trees of polygonal domains $(\Sigma _{v})_{{v\in V}}$ with each edge $e=(v,w)$ represented by two punctures $z_{e}^{-}\in \Sigma _{v}$ and $z_{e}^{+}\in \Sigma _{w}$. The thin neighbourhoods of these punctures are biholomorphic to half-strips, and a neighbourhood of a tree of polygonal domains is obtained by gluing the domains together at the pairs of strip-like ends represented by the edges.
All isotropy groups of this uncompactified moduli space ${\mathcal {M}}(x_{0};x_{1},\ldots ,x_{d})$ are trivial; that is any disk biholomorphism $\psi :D\to D$ that fixes $d+1\geq 1$ marked points $\psi (z_{i})=z_{i}$, and preserves a pseudoholomorphic map $u=u\circ \psi$ must be the identity $\psi ={{\rm {id}}}_{D}$.
Proof:
In case $d=2$ this follows directly from the marked points, since any Mobius transformation that fixes three points is the identity.
In case $d=1$ this requires both the stability and finite energy conditions: The group of biholomorphisms that fix two marked points - i.e. the biholomorphisms of the strip - are shifts by $\mathbb{R}$. On the other hand, any J-holomorphic map $u:\mathbb{R} \times [0,1]\to M$ has nonnegative energy density $u^{*}\omega ={\bigl (}|\partial _{s}u|^{2}+|\partial _{t}u|^{2}{\bigr )}ds\wedge dt$ with $|\partial _{s}u|=|\partial _{t}u|$. If we now had nontrivial isotropy, i.e. $u(\tau +s,t)=u(s,t)$ for some $\tau >0$ and a nonconstant map $u$, then there would exist $s_{0},t_{0}\in \mathbb{R} \times [0,1]$ with $|\partial _{s}u(s_{0},t_{0})|=|\partial _{t}u(s_{0},t_{0})|>0$ and thus $\textstyle \int _{{[s_{0}-{\frac {1}{2}}\tau ]\times [0,1]}}^{{[s_{0}+{\frac {1}{2}}\tau ]\times [0,1]}}u^{*}\omega >0$. However, this is in contradiction to $u$ having finite energy,
$\infty >\textstyle \int _{{\mathbb{R} \times [0,1]}}u^{*}\omega =\sum _{{k\in \mathbb{Z } }}\int _{{[s_{0}+(k-{\frac {1}{2}})\tau ]\times [0,1]}}^{{[s_{0}+(k+{\frac {1}{2}})\tau ]\times [0,1]}}u^{*}\omega =\sum _{{k\in \mathbb{Z } }}\int _{{[s_{0}-{\frac {1}{2}}\tau ]\times [0,1]}}^{{[s_{0}+{\frac {1}{2}}\tau ]\times [0,1]}}u^{*}\omega .$
Next, to construct the Gromov-compactified moduli spaces $\overline {\mathcal {M}}(x_{0};x_{1},\ldots ,x_{d})$ we have to add various strata to the moduli space of pseudoholomorphic polygons without breaking or nodes ${\mathcal {M}}(x_{0};x_{1},\ldots ,x_{d})$ defined above.
This is done precisely in the general construction below, but roughly requires to include breaking and bubbling, in particular
• include degenerate pseudoholomorphic polygons given by a tuple of pseudoholomorphic maps $u_{v}:\Sigma _{v}\to M$ whose domain is a nontrivial tree of domains $[(\Sigma _{v})_{{v\in V}},(z_{e}^{\pm })_{{e\in E}}]\in \overline {\mathcal {M}}_{{d+1}}$;
• allow for Floer breaking at each puncture of the domains $\Sigma _{v}$, i.e. a finite string of pseudoholomorphic strips in ${\mathcal {M}}(x;x'),{\mathcal {M}}(x';x''),\ldots ,{\mathcal {M}}(x^{{(k)}};x_{i})$;
• allow for disk bubbling at any boundary point of the above domains, i.e. a tree, each of whose vertices is represented by a pseudoholomorphic disk, with edges representing nodes - given by marked points on different disks at which the maps satisfy a matching condition;
• allow for sphere bubbling at any (boundary or interior) point of each of the above domains, i.e. a tree, each of whose vertices is represented by a pseudoholomorphic sphere, with edges representing nodes - given by marked points on different spheres at which the maps satisfy a matching condition.
We will see that sphere bubbling does not contribute to the boundary stratification of these moduli spaces, so that the boundary stratification and thus the algebraic structure arising from these moduli spaces is induced by Floer breaking and disk bubbling. (On the other hand, sphere bubbling will be the only source of nontrivial isotropy.) The boundary strata arising from Floer breaking are fiber products of other moduli spaces of pseudoholomorphic polygons over finite sets of Lagrangian intersection points, which indicates an algebraic composition in this finitely generated Floer chain complex.
Disk bubbling, on the other hand, in the present setting yields boundary strata that are fiber products over the Lagrangian submanifold specified by the boundary condition, which is problematic for a combination of algebra and regularity reasons.
The corresponding algebraic composition requires a push-pull construction on some space of chains, currents, or differential forms on the Lagrangian. However, such constructions require transversality of the chains to the evaluation maps from the regularized moduli spaces, so that a rigorous construction of the $A_{\infty }$-structure in this setting - as in the approach by Fukaya et al - requires a complicated infinite iteration.
We will resolve this issue as in [J.Li thesis] by following another earlier proposal by Fukaya-Oh to allow disks to flow apart along a Morse trajectory, thus yielding disk trees which are constructed next - still ignoring sphere bubbling.
## Pseudoholomorphic disk trees for a fixed Lagrangian
If the Lagrangians are all the same, $L_{0}=L_{1}=\ldots =L_{d}=:L$, then our construction is based on pseudoholomorphic disks
$u:D\to M,\qquad u(\partial D)\subset L,\qquad \overline \partial _{J}u=0.$
Such disks (modulo reparametrization by biholomorphisms of the disk) also arise from Gromov-compactifying other moduli spaces of pseudoholomorphic curves in which energy concentrates at a boundary point. To capture this bubbling algebraically, we work throughout with the Morse function $f:L\to \mathbb{R}$ chosen in the setup of the morphism space ${\text{Hom}}(L,L)=\textstyle \sum _{{x\in {\text{Crit}}(f)}}\Lambda x$. We also choose a metric on $L$ so that the gradient vector field $\nabla f$ satisfies the Morse-Smale conditions and an additional technical assumption in [1] which guarantees a natural smooth manifold-with-boundary-and-corners structure on the compactified Morse trajectory spaces $\overline {\mathcal {M}}(L,L),\overline {\mathcal {M}}(p^{-},L),\overline {\mathcal {M}}(L,p^{+}),\overline {\mathcal {M}}(p^{-},p^{+})$ for $p^{\pm }\in {\text{Crit}}(f)$. This smooth structure is essentially induced by the requirement that the evaluation maps at positive and negative ends ${{\rm {ev^{\pm }}}}:\overline {\mathcal {M}}(\ldots )\to L$ are smooth. With that data and the fixed almost complex structure $J$ we can construct the moduli spaces of pseudoholomorphic disk trees for any tuple $x_{0},x_{1},\ldots ,x_{d}\in {{\rm {Crit}}}(f)$ as in JL:
${\mathcal {M}}(x_{0};x_{1},\ldots ,x_{d}):={\bigl \{}(T,\underline {\gamma },\underline {z},\underline {u})\,{\big |}\,{\text{1. - 5.}}{\bigr \}}/\sim$
where
1. $T$ is an ordered tree with sets of vertices $V=V^{m}\cup V^{c}$ and edges $E$,
equipped with orientations towards the root, orderings of incoming edges, and a partition into main and critical (leaf and root) vertices as follows:
• The edges $E\subset V\times V\setminus \Delta _{V}$ are oriented towards the root vertex $v_{0}\in V$ of the tree, i.e. for $e=(v,w)\in E$ the outgoing vertex $w$ is still connected to the root after removing $e$. Thus each vertex $v\in V$ has a unique outgoing edge $e_{v}^{0}=(v,\;\cdot \;)\in E$ (except for the root vertex which has no outgoing edge) and a (possibly empty) set of incoming edges $E_{v}^{{{\rm {in}}}}=\{e=(\;\cdot \;,v)\in E\}$. Moreover, the set of incoming edges is ordered, $E_{v}^{{{\rm {in}}}}=\{e_{v}^{1},\ldots ,e_{v}^{{|v|-1}}\}$ with $|v|$ denoting the valence - number of attached edges - of $v$.
• The set of vertices is partitioned $V=V^{m}\sqcup V^{c}$ into the sets of main vertices $V^{m}$ and the set of critical vertices $V^{c}=\{v_{0}^{c},v_{1}^{c},\ldots v_{d}^{c}\}$. The latter is ordered to start with the root $v_{0}^{c}=v_{0}$, which is required to have a single edge $\{e_{{v_{0}}}^{1}\}=E_{{v_{0}}}^{{{\rm {in}}}}$, and then contains d leaves $v_{i}^{c}$ of the tree (i.e. with $E_{{v_{i}^{c}}}^{{{\rm {in}}}}=\emptyset$), with order induced by the orientation and order of the edges (with the root being the minimal vertex).
2. $\underline {\gamma }=(\underline {\gamma }_{e})_{{e\in E}}$ is a tuple of generalized Morse trajectories for each edge
in the following compactified Morse trajectory spaces:
• $\underline {\gamma }_{e}\in \overline {\mathcal {M}}(x_{i},x_{j})$ for any edge $e=(v_{i}^{c},v_{j}^{c})$ between critical vertices;
• $\underline {\gamma }_{e}\in \overline {\mathcal {M}}(x_{i},L)$ for any edge $e=(v_{i}^{c},w)$ from a critical vertex $v_{i}^{c}$ to a main vertex $w\in V^{m}$;
• $\underline {\gamma }_{e}\in \overline {\mathcal {M}}(L,x_{j})$ for any edge $e=(v,v_{j}^{c})$ from a main vertex $v\in V^{m}$ to a critical vertex $v_{j}^{c}$;
• $\underline {\gamma }_{e}\in \overline {\mathcal {M}}(L,L)$ for any edge $e=(v,w)$ between main vertices $v,w\in V^{m}$.
3. $\underline {z}=(\underline {z}_{v})_{{v\in V^{m}}}$ is a tuple of boundary marked points for each main vertex
that correspond to the edges of $T$ and are ordered counter-clockwise as follows:
• For each main vertex $v$ there are $|v|$ pairwise disjoint marked points $\underline {z}_{v}=(z_{e}^{v})_{{e\in \{e_{v}^{0}\}\cup E_{v}^{{{\rm {in}}}}}}\subset \partial D$ on the boundary of a disk.
• The order $\{e_{v}^{0}\}\cup E_{v}^{{{\rm {in}}}}=\{e_{v}^{0},e_{v}^{1},\ldots ,e_{v}^{{|v|-1}}\}$ of the edges corresponds to a counter-clockwise order of the marked points $z_{{e_{v}^{0}}}^{v},z_{{e_{v}^{1}}}^{v},\ldots ,z_{{e_{v}^{{|v|-1}}}}^{v}\in \partial D$.
• The marked points can also be denoted as $z_{e}^{-}=z_{e}^{v}$ and $z_{e}^{+}=z_{e}^{w}$ by the edges $e=(v,w)\in E$ for which $v\in V^{m}$ or $w\in V^{m}$.
4. $\underline {u}=(\underline {u}_{v})_{{v\in V^{m}}}$ is a tuple of pseudoholomorphic disks for each main vertex,
that is each $v\in V^{m}$ is labeled by a smooth map $u_{v}:D\to M$ satisfying Cauchy-Riemann equation, Lagrangian boundary condition, finite energy, and matching conditions as follows:
• The Cauchy-Riemann equation is $\overline \partial _{J}u_{v}=0$.
• The Lagrangian boundary condition is $u_{v}(\partial D)\subset L<\infty$.
• The finite energy condition is $\textstyle \int _{D}u_{v}^{*}\omega <\infty$.
• The pseudholomorphic disks can also be indexed as $u_{e}^{-}=u_{v}$ and $u_{e}^{+}=u_{w}$ by the edges $e=(v,w)\in E$ for which $v\in V^{m}$ or $w\in V^{m}$. In that notation, they satisfy the matching conditions with the generalized Morse trajectories $u_{e}^{\pm }(z_{e}^{\pm })={{\rm {ev}}}^{\pm }(\underline {\gamma }_{e})$ whenever $v_{e}^{\pm }\in V^{m}$.
5. The disk tree is stable
in the sense that
any main vertex $v\in V^{m}$ whose disk has zero energy $\textstyle \int u_{v}^{*}\omega =0$ (which is equivalent to $u_{v}$ being constant) has valence $|v|\geq 3$.
Finally, two pseudoholomorphic disk trees are equivalent $(T,\underline {\gamma },\underline {z},\underline {u})\sim (T',\underline {\gamma }',\underline {z}',\underline {u}')$ if
there is a tree isomorphism $\zeta :T\to T'$ and a tuple of disk biholomorphisms $(\psi _{v}:D\to D)_{{v\in V^{m}}}$ which preserve the tree, Morse trajectories, marked points, and pseudoholomorphic curves in the sense that
• $\zeta$ preserves the tree structure and order of edges;
• $\underline {\gamma }_{e}=\underline {\gamma }'_{{\zeta (e)}}$ for every $e\in E$;
• $\psi _{v}(z_{e}^{v})={z'}_{{\zeta (e)}}^{{\zeta (v)}}$ for every $v\in V^{m}$ and adjacent edge $e\in E_{v}$;
• the pseudoholomorphic disks are related by reparametrization, $u_{v}=u'_{{\zeta (v)}}\circ \psi _{v}$ for every $v\in V^{m}$.
The domains of the disk trees are never stable for $d=0,1$, but need to be studied to construct the differential $\mu ^{1}$ on the Floer chain complex and the curvature term $\mu ^{0}$ that may obstruct $\mu ^{1}\circ \mu ^{1}=0$. For $d\geq 2$ the domains of the disk trees represent elements in a Deligne-Mumford space as follows:
Any equivalence class of disk trees $[(T,\underline {\gamma },\underline {z},\underline {u})]$ induces a domain tree $[(T',\underline {\ell },\underline {z})]$
by forgetting the Morse trajectories and pseudoholomorphic maps as follows:
1. The tree $T'$ is obtained from $T$ by replacing critical vertices and their outgoing edges by incoming semi-infinite edges of the new tree $T'$. We also replace the critical root vertex and its incoming edge by an outgoing semi-infinite edge of the new tree $T'$. The new tree $T'$ retains the orientations of edges and inherits an order of the edges from $T$. Its root is the unique main vertex from which there was an edge to the critical root vertex in $T$.
2. Every vertex $v\in V'=V^{m}$ of $T'$ then represents a disk domain $D_{v}=D$.
3. Every edge $e\in E'=E$ is labeled with the length $\ell _{e}:={\tfrac {\ell (\underline {\gamma }_{e})}{1-\ell (\underline {\gamma }_{e})}}\in [0,\infty ]$ of the associated generalized Morse trajectory (recalling that the function $\ell :\overline {{\mathcal {M}}}(\ldots )\to [0,1]$ on compactified Morse trajectory spaces is the renormalized length). For the semi-infinite edges, this length is automatically $\ell _{e}=\infty$ since the associated Morse trajectories are semi-infinite.
4. The domain for each vertex $v\in V'$ is marked by $|v|$ boundary points $\underline {z}_{v}=(z_{e}^{v})_{{e\in \{e_{v}^{0}\}\cup E_{v}^{{{\rm {in}}}}}}\subset \partial D_{v}$, ordered counter-clockwise.
5. Two such trees are equivalent $[(T,\underline {\ell },\underline {z})]\sim [(T',\underline {\ell }',\underline {z}')]$ if there is a tree isomorphism $\zeta :T\to T'$ and a tuple of disk biholomorphisms $(\psi _{v}:D\to D)_{{v\in V^{m}}}$ such that $\zeta$ preserves the ordered tree structure and lengths $\ell _{e}=\ell '_{{\zeta (e)}}$ for every $e\in E$, and the marked points are preserved $\psi _{v}(z_{e}^{v})=z{'\;\zeta (v)}_{{\zeta (e)}}$ for every $v\in V^{m}$ and adjacent $e\in E$.
For $d\geq 2$, such a domain tree is called stable if every vertex has valence $|v|\geq 3$ - i.e. there are at least three marked points on each disk $D_{v}$. The domain trees for $d=0,1$ are never stable, but both cases need to be included in our moduli space constructions: The differential $\mu ^{1}$ on the Floer chain complex $Hom(L,L)=\textstyle \sum _{{x\in {{\rm {Crit}}}(f)}}\Lambda \;x$ is constructed by counting the elements of ${\mathcal {M}}(x_{0};x_{1})$. The curvature term $\mu ^{0}$, which is constructed from moduli spaces ${\mathcal {M}}(x_{0})$ with no incoming critical points, serves to algebraically encode disk bubbling in any moduli space involving a Lagrangian boundary condition on $L$.
For $d\geq 2$, while the above trees are not necessarily stable, they induce unique stable rooted metric ribbon trees $(T,\underline {\ell })$ in the sense of [Def.2.7, MW], by forgetting the marked points, forgetting every leaf of valence 1 and its outgoing edge, and replacing every vertex $v$ of valence 2 and its incoming and outgoing edges $(v^{-},v),(v,v^{+})$ by a single edge $(v^{-},v^{+})$ of length $\ell _{{(v^{-},v^{+})}}=\ell _{{(v^{-},v)}}+\ell _{{(v,v^{+})}}$. The space of such stable rooted metric ribbon trees - where a tree containing an edge of length $\ell _{e}=0$ is identified with the tree in which this edge and its adjacent vertices are replaced by a single vertex - is another topological representation of the Deligne Mumford space $\overline {\mathcal {M}}_{{d+1}}$, as discussed in [BV]. Its boundary strata are given by trees with interior edges of length $\ell _{e}=\infty$.
We now expect the boundary stratification of the moduli spaces of disk trees ${\mathcal {M}}(x_{0};x_{1},\ldots ,x_{d})$ - if/once regular - to arise exclusively from breaking of the Morse trajectories representing edges of the disk trees. This is made rigorous in [J.Li thesis] under the assumption that the almost complex structure $J$ can be chosen such that there exist no nonconstant $J$-holomorphic spheres in the symplectic manifold $M$. In that special case, all isotropy groups are trivial by [Prop.2.5, J.Li thesis]; that is any equivalence between a disk tree and itself, $(T,\underline {\gamma },\underline {z},\underline {u})\sim (T,\underline {\gamma },\underline {z},\underline {u})$, is given by the trivial tree isomorphism $\zeta :T\to T$, and the only disk biholomorphisms $(\psi _{v}:D\to D)_{{v\in V^{m}}}$ which preserve the marked points and pseudoholomorphic disk maps are the identity maps $\psi _{v}={{\rm {id}}}_{D}$. In this case, the moduli spaces of disk trees ${\mathcal {M}}(x_{0};x_{1},\ldots ,x_{d})$ will moreover be Gromov-compact (with respect to the Gromov topology) since sphere bubbling is ruled out and disk bubbling is captured by edges labeled with constant, zero length, Morse trajectories.
In general, we will Gromov-compactify ${\mathcal {M}}(x_{0};x_{1},\ldots ,x_{d})$ in the general construction below by allowing for sphere bubble trees (which we formalize next) to develop at any point of the disk and polygon domains. These will also be a source of generally nontrivial isotropy.
## Sphere bubble trees
The sphere bubble trees that are relevant to the compactification of the moduli spaces of pseudoholomorphic polygons are genus zero stable maps with one marked point, as described in e.g. [Chapter 5, McDuff-Salamon]. For a fixed almost complex structure $J$, we can use the combinatorial simplification of working with a single marked point to construct the moduli space of sphere bubble trees as
$\overline {\mathcal {M}}_{{0,1}}(J):={\bigl \{}(T,\underline {z},\underline {u})\,{\big |}\,{\text{1. - 4.}}{\bigr \}}/\sim$
where
1. $T$ is a tree with sets of vertices $V$ and edges $E$, and a distinguished root vertex $v_{0}\in V$, which we use to orient all edges towards the root.
2. $\underline {z}=(\underline {z}_{v})_{{v\in V}}$ is a tuple of marked points on the spherical domains $\Sigma ^{v}=S^{2}$,
indexed by the edges of $T$, and including a special root marked point as follows:
• For each vertex $v\neq v_{0}$ the tuple of mutually disjoint marked points $\underline {z}_{v}=(z_{e}^{v})_{{e\in E_{v}}}\subset S^{2}$ is indexed by the edges $E_{v}=\{e\in E\,|\,e=(v,\,\cdot \,)\;{\text{or}}\;e=(\,\cdot \,,v)\}$ adjacent to $v$.
• For the root vertex $v_{0}$ the tuple of mutually disjoint marked points $\underline {z}_{v}=(z_{e}^{v})_{{e\in E_{{v_{0}}}}}\subset S^{2}$ is also indexed by the edges adjacent to $v_{0}$, but is also required to be disjoint from the fixed marked point $z_{0}=0\in S^{2}\simeq \mathbb{C} \cup \{\infty \}$.
• The marked points, except for $z_{0}$, can also be denoted as $z_{e}^{-}=z_{e}^{v}$ and $z_{e}^{+}=z_{e}^{w}$ by the edges $e=(v,w)\in E$.
3. $\underline {u}=(\underline {u}_{v})_{{v\in V}}$ is a tuple of pseudoholomorphic spheres for each vertex,
that is each $v\in V$ is labeled by a smooth map $u_{v}:S^{2}\to M$ satisfying Cauchy-Riemann equation, finite energy, and matching conditions as follows:
• The Cauchy-Riemann equation is $\overline \partial _{J}u_{v}=0$.
• The finite energy condition is $\textstyle \int _{{S^{2}}}u_{v}^{*}\omega <\infty$.
• The matching conditions are $u^{v}(z_{e}^{v})=u^{w}(z_{e}^{w})$ for each edge $e=(v,w)\in E$.
4. The sphere bubble tree is stable
in the sense that
any vertex $v\in V$ whose map has zero energy $\textstyle \int u_{v}^{*}\omega =0$ (which is equivalent to $u_{v}$ being constant) has valence $|v|\geq 3$. Here the marked point $z_{0}$ counts as one towards the valence $|v_{0}|$ of the root vertex; in other words the root vertex can be constant with just two adjacent edges.
Finally, two sphere bubble trees are equivalent $(T,\underline {z},\underline {u})\sim (T',\underline {z}',\underline {u}')$ if
there is a tree isomorphism $\zeta :T\to T'$ and a tuple of sphere biholomorphisms $(\psi _{v}:S^{2}\to S^{2})_{{v\in V}}$ which preserve the tree, marked points, and pseudoholomorphic curves in the sense that
• $\zeta$ preserves the tree structure, in particular maps the root $v_{0}$ to the root $v_{0}'$;
• $\psi _{{v_{0}}}(0)=0$ and $\psi _{v}(z_{e}^{v})={z'}_{{\zeta (e)}}^{{\zeta (v)}}$ for every $v\in V$ and adjacent edge $e\in E_{v}$;
• the pseudoholomorphic spheres are related by reparametrization, $u_{v}=u'_{{\zeta (v)}}\circ \psi _{v}$ for every $v\in V$.
To attach such sphere bubble trees to the generalized pseudoholomorphic polygons below, we will use the evaluation map (which is well defined independent of the choice of representative)
${\text{ev}}_{0}\,:\;\overline {\mathcal {M}}_{{0,1}}(J)\;\to \;M,\qquad {\bigl [}T,\underline {z},\underline {u}{\bigr ]}\;\mapsto \;u_{{v_{0}}}(0).$
We will moreover make use of the symplectic area function
$\omega \,:\;\overline {\mathcal {M}}_{{0,1}}(J)\;\to \;\mathbb{R} \qquad {\bigl [}T,\underline {z},\underline {u}{\bigr ]}\;\mapsto \;\textstyle \sum _{{v\in V}}\int _{{S^{2}}}u_{v}^{*}\omega \;=\;\langle [\omega ],\sum _{{v\in V}}(u_{v})_{*}[S^{2}]\rangle ,$
which only depends on the total homology class of the sphere bubble tree $\beta ={\bigl [}T,\underline {z},\underline {u}{\bigr ]}$,
$\textstyle [\beta ]:=\sum _{{v\in V}}(u_{v})_{*}[S^{2}]\;\in \;H_{2}(M).$
## General moduli space of pseudoholomorphic polygons
For the construction of a general $A_{\infty }$-composition map we are given $d+1\geq 1$ Lagrangians $L_{0},\ldots ,L_{d}\subset M$ and a fixed autonomous Hamiltonian function $H_{{L_{i},L_{j}}}:M\to \mathbb{R}$ for each pair $L_{i}\neq L_{j}$ whose time-1 flow provides transverse intersections $\phi _{{L_{i},L_{j}}}(L_{i})\pitchfork L_{j}$. To simplify notation for consecutive Lagrangians in the list, we index it cyclically by $i\in \mathbb{Z } _{{d+1}}$ and abbreviate $\phi _{i}:=\phi _{{L_{{i-1}},L_{i}}}$ so that we have $\phi _{i}(L_{{i-1}})\pitchfork L_{i}$ whenever $L_{{i-1}}\neq L_{i}$, and in particular $\phi _{0}(L_{d})\pitchfork L_{0}$ unless $L_{d}=L_{0}$. Now, given generators $x_{0}\in {\text{Crit}}(L_{0},L_{d}),$ $x_{1}\in {\text{Crit}}(L_{0},L_{1}),\ldots ,$ $x_{d}\in {\text{Crit}}(L_{{d-1}},L_{d})$ of these morphism spaces, we construct the Gromov-compactified moduli space of generalized pseudoholomorphic polygons by combining the two special cases above with sphere bubble trees,
${\mathcal {M}}(x_{0};x_{1},\ldots ,x_{d}):={\bigl \{}(T,\underline {\gamma },\underline {z},\underline {w},\underline {\beta },\underline {u})\,{\big |}\,{\text{1. - 8.}}{\bigr \}}/\sim$
where
1. $T$ is an ordered tree with sets of vertices $V=V^{m}\cup V^{c}$ and edges $E$,
equipped with orientations towards the root, orderings of incoming edges, and a partition into main and critical (leaf and root) vertices as follows:
• The edges $E\subset V\times V\setminus \Delta _{V}$ are oriented towards the root vertex $v_{0}\in V$ of the tree, so that each vertex $v\in V$ has a unique outgoing edge $e_{v}^{0}=(v,\;\cdot \;)\in E$ (except for the root vertex which has no outgoing edge) and a (possibly empty) set of incoming edges $E_{v}^{{{\rm {in}}}}=\{e=(\;\cdot \;,v)\in E\}$.
• The set of incoming edges is ordered, $E_{v}^{{{\rm {in}}}}=\{e_{v}^{1},\ldots ,e_{v}^{{|v|-1}}\}$. This induces a cyclic order on the set of all edges $E_{v}:=\{e_{v}^{0},e_{v}^{1},\ldots ,e_{v}^{{|v|-1}}\}$ adjacent to $v$, by setting $e_{v}^{{|v|}}=e_{v}^{0}$, and we will denote consecutive edges in this order by $e=e_{v}^{i},e+1=e_{v}^{{i+1}}$. In particular this yields $e_{v}^{0}+i=e_{v}^{i}$.
• The set of vertices is partitioned $V=V^{m}\sqcup V^{c}$ into the sets of main vertices $V^{m}$ and critical vertices $V^{c}=\{v_{0}^{c},v_{1}^{c},\ldots v_{d}^{c}\}$. The latter is ordered to start with the root $v_{0}^{c}=v_{0}$ and then contains d leaves $v_{i}^{c}$ of the tree, with order induced by the orientation and order of the edges.
• The root vertex $v_{0}^{c}\in V^{c}$ has a single edge $\{e_{{v_{0}}}^{1}=(v,v_{0}^{c})\}=E_{{v_{0}}}^{{{\rm {in}}}}=E_{{v_{0}}}$, and this attaches to a main vertex $v\in V^{m}$ except for one special case: For $d=1$ and $L_{d}=L_{0}$ we allow the tree with a single edge $e=(v_{1}^{c},v_{0}^{c})$ between its two critical vertices $V=\{v_{0}^{c},v_{1}^{c}\}$.
2. The tree structure induces tuples of Lagrangians $\underline {L}=(\underline {L}^{v})_{{v\in V^{m}}}$
that label the boundary components of domains in overall counter-clockwise order $L_{0},\ldots ,L_{d}$ as follows:
• For each main vertex $v\in V^{m}$ the Lagrangian label $\underline {L}^{v}=(L_{e}^{v})_{{e\in E_{v}}}$ is a cyclic sequence of Lagrangians $L_{e}^{v}\in \{L_{0},\ldots ,L_{d}\}$ indexed by the adjacent edges $E_{v}$ (which will become the boundary condition on $(\partial \Sigma ^{v})_{e}$).
• For each edge $e=(v^{-},v^{+})\in E$ the Lagrangian labels satisfy a matching condition as follows:
• The edge from a critical leaf $v^{-}=v_{i}^{c}\in V^{c}$ requires $L_{e}^{{v^{+}}}=L_{i},L_{{e-1}}^{{v^{+}}}=L_{{i-1}}$.
• The edge to the critical root $v^{+}=v_{0}^{c}\in V^{c}$ requires $L_{e}^{{v^{-}}}=L_{0},L_{{e-1}}^{{v^{-}}}=L_{d}$.
• Any edge between main vertices $v^{-},v^{+}\in V^{m}$ requires $L_{{e}}^{{v^{-}}}=L_{{e-1}}^{{v^{+}}}$ and $L_{{e-1}}^{{v^{-}}}=L_{{e}}^{{v^{+}}}$.
• Since $T$ has no further leaves, this determines the Lagrangian labels uniquely.
3. $\underline {\gamma }=(\underline {\gamma }_{e})_{{e\in E}}$ is a tuple of generalized Morse trajectories
in the following compactified Morse trajectory spaces:
• Any edge $e=(v_{i}^{c},w)$ from a critical leaf $v_{i}^{c}$ to a main vertex $w\in V^{m}$ is labeled by a half-infinite Morse trajectory $\underline {\gamma }_{e}\in \overline {\mathcal {M}}(x_{i},L_{i})$ if $L_{{i-1}}=L_{i}$, resp. by the constant $\underline {\gamma }_{e}\equiv x_{i}\in {{\rm {Crit}}}(L_{{i-1}},L_{i})$ in the discrete space $\phi _{i}(L_{{i-1}})\cap L_{i}$ if $L_{{i-1}}\neq L_{i}$.
• If the edge to the root $e=(v,v_{0}^{c})$ attaches to a main vertex $v\in V^{m}$ then it is labeled by a half-infinite Morse trajectory $\underline {\gamma }_{e}\in \overline {\mathcal {M}}(L_{0},x_{0})$ if $L_{d}=L_{0}$, resp. by the constant $\underline {\gamma }_{e}\equiv x_{0}\in {{\rm {Crit}}}(L_{d},L_{0})$ in the discrete space $\phi _{0}(L_{d})\cap L_{0}$ if $L_{d}\neq L_{0}$.
• An edge $e=(v_{i}^{c},v_{j}^{c})$ between critical vertices is labeled by an infinite Morse trajectory $\underline {\gamma }_{e}\in \overline {\mathcal {M}}(x_{i},x_{j})$ (this occurs only for $d=1$ with $L_{0}=L_{1}$ and the tree with one edge $e=(v_{1}^{c},v_{0}^{c})$).
• Any edge $e=(v,w)$ between main vertices $v,w\in V^{m}$ is labeled by a finite or infinite Morse trajectory $\underline {\gamma }_{e}\in \overline {\mathcal {M}}(L_{e}^{v},L_{e}^{v})$ in case $L_{e}^{v}=L_{{e-1}}^{v}$, resp. by a constant $\underline {\gamma }_{e}\equiv x_{e}\in {{\rm {Crit}}}(L_{{e-1}}^{v},L_{{e}}^{v})$ in the discrete space $\phi _{{L_{{e-1}}^{v},L_{{e}}^{v}}}(L_{{e-1}}^{v})\cap L_{e}^{v}$ in case $L_{e}^{v}\neq L_{{e-1}}^{w}$. (Recall the matching condition $L_{e}^{v}=L_{{e-1}}^{w}$ and $L_{{e-1}}^{v}=L_{e}^{w}$ from 2.)
4. $\underline {z}=(\underline {z}_{v})_{{v\in V^{m}}}$ is a tuple of boundary points
that correspond to the edges of $T$, are ordered counter-clockwise, and associate complex domains $\Sigma ^{v}:=D\setminus \underline {z}_{v}$ to the vertices as follows:
• For each main vertex $v$ there are $|v|$ pairwise disjoint marked points $\underline {z}_{v}=(z_{e}^{v})_{{e\in E_{v}}}\subset \partial D$ on the boundary of a disk.
• The order $E_{v}=\{e_{v}^{0},e_{v}^{1},\ldots ,e_{v}^{{|v|-1}}\}$ of the edges corresponds to a counter-clockwise order of the marked points $z_{{e_{v}^{0}}}^{v},z_{{e_{v}^{1}}}^{v},\ldots ,z_{{e_{v}^{{|v|-1}}}}^{v}\in \partial D$.
• The marked points can also be denoted as $z_{e}^{-}=z_{e}^{v}$ and $z_{e}^{+}=z_{e}^{w}$ by the edges $e=(v,w)\in E$ for which $v\in V^{m}$ or $w\in V^{m}$
• To each main vertex $v\in V^{m}$ we associate the punctured disk $\Sigma ^{v}:=D\setminus \underline {z}_{v}$. Then the marked points $\underline {z}^{v}=(z_{e}^{v})_{{e\in E_{v}}}\subset \partial D$ partition the boundary into $|v|$ connected components $\partial \Sigma ^{v}=\textstyle \sqcup _{{e\in E_{v}}}(\partial \Sigma ^{v})_{e}$ such that the closure of each component $(\partial \Sigma ^{v})_{e}$ contains the marked points $z_{e}^{v},z_{{e+1}}^{v}$.
5. $\underline {w}=(\underline {w}_{v})_{{v\in V^{m}}}$ is a tuple of sphere bubble tree attaching points for each main vertex $v\in V^{m}$, given by an unordered subset $\underline {w}_{v}\subset \Sigma ^{v}\setminus \partial \Sigma ^{v}$ of the interior of the domain.
6. $\underline {\beta }=(\beta _{w})_{{w\in \underline {w}}}\subset \overline {\mathcal {M}}_{{0,1}}(J)$ is a tuple of sphere bubble trees $\beta _{w}\in \overline {\mathcal {M}}_{{0,1}}(J)$ indexed by the disjoint union $\underline {w}=\textstyle \bigsqcup _{{v\in V^{m}}}\underline {w}_{v}$ of sphere bubble tree attaching points.
7. $\underline {u}=(\underline {u}_{v})_{{v\in V^{m}}}$ is a tuple of pseudoholomorphic maps for each main vertex,
that is each $v\in V^{m}$ is labeled by a smooth map $u_{v}:\Sigma ^{v}\to M$ satisfying Cauchy-Riemann equation, Lagrangian boundary conditions, finite energy, and matching conditions as follows:
• The Cauchy-Riemann equation is
$0=\overline \partial _{{J,Y}}u_{v}:={\bigl (}{{\rm {d}}}u_{v}+Y_{v}\circ u_{v}{\bigr )}^{{0,1}}={\tfrac 12}{\bigl (}J_{v}(u_{v})\circ ({{\rm {d}}}u_{v}-Y_{v}(\cdot ,u_{v}))-({{\rm {d}}}u_{v}-Y_{v}(\cdot ,u_{v}))\circ i{\bigr )}.$
Here $Y_{v}:{{\rm {T}}}^{*}\Sigma ^{v}\times M\to {{\rm {T}}}M$ is a vector-field-valued 1-form on $\Sigma ^{v}$ that is chosen compatibly with the fixed Hamiltonian perturbations as follows:
On the thin part $\iota _{e}^{v}:[0,\infty )\times [0,1]\hookrightarrow \Sigma ^{v}$ near each puncture $z_{e}^{v}$ we have $(\iota _{e}^{v})^{*}Y_{v}=X_{{L_{{e-1}}^{v},L_{e}^{v}}}\,{{\rm {d}}}t$.
In particular, this convention together with our symmetric choice of Hamiltonian perturbations $X_{{L_{i},L_{j}}}=-X_{{L_{j},L_{i}}}$ forces the vector-field-valued 1-form on $\Sigma ^{v}\simeq \mathbb{R} \times [0,1]$ in case $|v|=2$ to be $\mathbb{R}$-invariant, $Y_{v}=X_{{L_{0},L_{1}}}\,{{\rm {d}}}t$ if $L_{i}$ are the Lagrangian labels for the boundary components $\mathbb{R} \times \{i\}$.
Here and in the following we denote $X_{{L_{i},L_{j}}}:=0$ in case $L_{i}=L_{j}$, so that $(\iota _{e}^{v})^{*}Y_{v}=0$ in case $L_{{e-1}}^{v}=L_{e}^{v}$.
The Hamiltonian perturbations $Y_{v}$ should be cut off to vanish outside of the thin parts of the domains $\Sigma _{v}$. However, there may be thin parts of a surface $\Sigma _{v}$ that are not neighborhoods of a puncture. On these, we must choose the Hamiltonian-vector-field-valued one-form $Y_{v}$ compatible with gluing as in [Seidel book]. For example, in the neighbourhood of a tree with an edge $e=(v,w)$ between main vertices, there are trees in which this edge is removed, the two vertices are replaced by a single vertex $v\#w$, and the surfaces $\Sigma _{v},\Sigma _{w}$ are replaced by a single glued surface $\Sigma _{{v\#w}}=\Sigma _{v}\#_{R}\Sigma _{w}$ for $R\gg 1$. Compatibility with gluing requires that the Hamiltonian perturbation on these glued surfaces is also given by a gluing construction for Hamiltonians $Y_{{v\#w}}=Y_{v}\#_{R}Y_{w}$ (in which the two perturbations $Y_{v},Y_{w}$ agree and hence can be matched over a long neck $[-R,R]\times [0,1]\subset \Sigma _{{v\#w}}$).
• The Lagrangian boundary conditions are $u_{v}(\partial \Sigma ^{v})\subset \underline {L}^{v}$; more precisely this requires $u_{v}{\bigl (}(\partial \Sigma ^{v})_{e}{\bigr )}\subset L_{e}^{v}$ for each adjacent edge $e\in E_{v}$.
• The finite energy condition is $\textstyle \int _{{\Sigma ^{v}}}u_{v}^{*}\omega <\infty$.
• The matching conditions for sphere bubble trees are $u^{v}(w)={\text{ev}}_{0}(\beta _{w})$ for each main vertex $v\in V^{m}$ and sphere bubble tree attaching point $w\in \underline {w}_{v}$.
• Finite energy together with the (perturbed) Cauchy-Riemann equation implies uniform convergence of $u_{v}$ near each puncture $z_{e}^{v}$, and the limits are required to satisfy the following matching conditions:
• For edges $e\in E_{v}$ whose Lagrangian boundary conditions $L_{{e-1}}^{v}=L_{e}^{v}$ agree, the map $u_{v}$ extends smoothly to the puncture $z_{e}^{v}$, and its value is required to match with the evaluation of the Morse trajectory $\underline {\gamma }_{e}$ associated to the edge $e=(v_{e}^{-},v_{e}^{+})$, that is $u_{v}(z_{e}^{v})={{\rm {ev}}}^{\pm }(\underline {\gamma }_{e})$ for $v=v_{e}^{\mp }$.
• For edges $e\in E_{v}$ with different Lagrangian boundary conditions $L_{{e-1}}^{v}\neq L_{e}^{v}$, the map $u_{e}^{v}:=(\iota _{e}^{v})^{*}u_{v}:[0,\infty )\times [0,1]\to M$ has a uniform limit $\lim _{{s\to \infty }}u^{v}(s,t)=\phi _{{L_{{e-1}}^{v},L_{e}^{v}}}^{{t-1}}(x_{e})$ for some $x_{e}\in \phi _{{L_{{e-1}}^{v},L_{e}^{v}}}(L_{{e-1}}^{v})\cap L_{e}^{v}={\text{Crit}}(L_{{e-1}}^{v},L_{e}^{v})$, and this limit intersection point is required to match with the value of the constant 'Morse trajectory' $\underline {\gamma }_{e}\in {\text{Crit}}(L_{{e-1}}^{v},L_{e}^{v})$ associated to the edge $e=(v_{e}^{-},v_{e}^{+})$, that is $x_{e}=\lim _{{s\to \infty }}u^{v}(s,1)=\underline {\gamma }_{e}$.
8. The generalized pseudoholomorphic polygon is stable
in the sense that
for any main vertex $v\in V^{m}$ with fewer than three special points $\#\underline {z}_{v}+2\#\underline {w}_{v}<3$, the map differential ${{\rm {d}}}_{z}u_{v}:{{\rm {T}}}_{z}\Sigma ^{v}\to {{\rm {T}}}_{{u_{v}(z)}}M$ must be injective on an open subset of $\Sigma ^{v}$.
Finally, two generalized pseudoholomorphic polygons are equivalent $(T,\underline {\gamma },\underline {z},\underline {w},\underline {\beta },\underline {u})\sim (T',\underline {\gamma }',\underline {z}',\underline {w}',\underline {\beta }',\underline {u}')$ if
there is a tree isomorphism $\zeta :T\to T'$ and a tuple of disk biholomorphisms $(\psi _{v}:D\to D)_{{v\in V^{m}}}$ which preserve the tree, Morse trajectories, marked points, and pseudoholomorphic curves in the sense that
• $\zeta$ preserves the tree structure and order of edges;
• $\underline {\gamma }_{e}=\underline {\gamma }'_{{\zeta (e)}}$ for every $e\in E$;
• $\psi _{v}(z_{e}^{v})={z'}_{{\zeta (e)}}^{{\zeta (v)}}$ for every $v\in V^{m}$ and adjacent edge $e\in E_{v}$;
• $\psi _{v}(\underline {w}_{v})=\underline {w}'_{{\zeta (v)}}$ for every $v\in V^{m}$;
• $\beta _{w}=\beta '_{{\psi _{v}(w)}}$ for every $v\in V^{m}$ and $w\in \underline {w}_{v}$;
• the pseudoholomorphic maps are related by reparametrization, $u_{v}=u'_{{\zeta (v)}}\circ \psi _{v}$ for every $v\in V^{m}$.
Warning: Our directional conventions differ somewhat from [Seidel book] and [J.Li thesis] as follows:
Unlike both references, we orient edges towards the root, in order to obtain a more natural interpretation of the leaves as incoming vertices as in [J.Li thesis], but unlike [Seidel book] which uses the language of 1 incoming striplike end and $d\geq 1$ outgoing striplike ends. Since we also insist on ordering the marked points counter-clockwise on the boundary of the disk, we then have to work with positive striplike ends $[0,\infty )\times [0,1]\hookrightarrow \Sigma ^{v}$ near each marked point $z_{e}^{v}$ for an incoming edge $e\in E_{v}^{{{\rm {in}}}}$ to make sure that the boundary components are labeled in order: $[0,\infty )\times \{0\}$ with $L_{{e-1}}^{v}$, and $[0,\infty )\times \{1\}$ with $L_{e}^{v}$. Analogously, a negative striplike end $(-\infty ,0]\times [0,1]\hookrightarrow \Sigma ^{v}$ near the marked point $z_{{e_{v}^{0}}}^{v}$ for the outgoing edge labels $[0,\infty )\times \{0\}$ with $L_{{e_{v}^{0}}}^{v}$ and $[0,\infty )\times \{1\}$ with $L_{{e_{v}^{{|v|-1}}}}^{v}$.
This amounts to working on Floer cohomology in the sense that for e.g. $L_{0}\pitchfork L_{1}$ the output of the differential $\mu ^{1}(x_{1})=\textstyle \sum _{{x_{0}\in {{\rm {Crit}}}(L_{0},L_{1})}}\sum _{{b\in {\mathcal {M}}^{0}(x_{0};x_{1})}}w(b)T^{{\omega (b)}}x_{0}$ includes a sum over (amongst other more complicated trees) pseudoholomorphic strips $b=[u:\mathbb{R} \times [0,1]\to M]$ with fixed positive limit $\lim _{{s\to \infty }}u(s,t)=x_{1}$.
If no Hamiltonian perturbations are involved in the construction of a moduli space of generalized pseudoholomorphic polygons - i.e. if for each pair $\{i,j\}\subset \{0,\ldots ,d\}$ the Lagrangians are either identical $L_{i}=L_{j}$ or transverse $L_{i}\pitchfork L_{j}$ - then the symplectic area function on the moduli space is defined by
$\omega :\overline {\mathcal {M}}(x_{0};x_{1},\ldots ,x_{d})\to \mathbb{R} ,\quad b={\bigl [}T,\underline {\gamma },\underline {z},\underline {w},\underline {\beta },\underline {u}{\bigr ]}\mapsto \omega (b):=\sum _{{v\in V^{m}}}\textstyle \int _{{\Sigma _{v}}}u_{v}^{*}\omega \;+\;\sum _{{\beta _{w}\in \underline {\beta }}}\omega (\beta _{w})=\langle [\omega ],[b]\rangle ,$
which - since $\omega |_{{L_{i}}}\equiv 0$ only depends on the total homology class of the generalized polygon
$[b]:=\sum _{{v\in V}}(\overline {u}_{v})_{*}[D]+\sum _{{\beta _{w}\in \underline {\beta }}}[\beta _{w}]\;\in \;H_{2}(M;L_{0}\cup L_{1}\ldots \cup L_{d}).$
Here $\overline {u}_{v}:D\to M$ is defined by unique continuous continuation to the punctures $z_{e}^{v}$ at which $L_{{e-1}}^{v}=L_{e}^{v}$ or $L_{{e-1}}^{v}\pitchfork L_{e}^{v}$.
Differential Geometric TODO:
In the presence of Hamiltonian perturbations, the definition of the symplectic area function needs to be adjusted to match with the symplectic action functional on the Floer complexes and satisfy some properties (which also need to be proven in the unperturbed case):
• $\omega (b)$ is invariant under deformations with fixed limits (used in proof of A-infty relations and invariance);
• a bound on $\omega (b)$ needs to imply Gromov-compactness ... which requires an area-energy identity for J-curves, but we are allowed (bounded!) error terms from e.g. Hamiltonian perturbations;
• invariance proofs arguing with 'upper triangular form' require contributions to $\mu ^{1}$ to be of positive symplectic area, or constant strips/disks for zero symplectic area;
• to work with the Novikov ring, rather than field, the symplectic area needs to be nonnegative for all polygons (not just the strips and other contributions to $\mu ^{1}$ for which this is automatic). It *might* be possible to achieve this by remembering that the Hamilton functions for each pair of Lagrangians are only fixed up to a constant (not see in the Hamiltonian vector field or time-1-flow), so when constructing the Hamiltonian perturbation vector fields over Deligne-Mumford spaces, it might be possible to shift e.g. the outgoing Hamiltonian in such a way that the vector field can be constructed with 'curvature terms of the correct sign'.
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2019-02-19 21:57:19
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http://math.stackexchange.com/questions/222629/why-does-mathfrako-l-cong-mathfrako-kn-imply-that-mathfrako-l-pi
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# Why does $\mathfrak{o}_L \cong \mathfrak{o}_K^n$ imply that $\mathfrak{o}_L/\pi_k\mathfrak{o}_L$ is a $k_K$ vector space of dimension $n$?
I'm trying to read a proof of the following proposition:
Let $L/K$ be a finite, separable extension of a complete discretely valued field. Then $e(L/K)f(L/K)=[L:K]$.
I'm stuck on the step:
We know $\mathfrak{o}_l \cong \mathfrak{o}_K^n$ as $\mathfrak{o}_K$ modules. So $\mathfrak{o}_L/\pi_K\mathfrak{o}_L$ is a $k_K$-vector space of dimension $n$
where $\pi_K$ is a uniformiser for $K$ (a generator of the maximal ideal of $\mathfrak{o}_K$) and $k_K=\mathfrak{o}_K/(\pi_K\mathfrak{o}_K$.
I see that the dimension must be at most $n$, since the images of the generators of $\mathfrak{o}_L$ in $\mathfrak{o}_L/\pi_K\mathfrak{o}_L$ span $\mathfrak{o}_L/\pi_K\mathfrak{o}_L$, but I don't see why they are still linearly independent.
-
Consider the exact sequence $$0 \to \pi_K\mathfrak{o}_L \cong \pi_K\mathfrak{o}_K^n \to \mathfrak{o}_K^n \to \mathfrak{o}_K^n/\pi_K \mathfrak{o}_K^n \to 0$$
Elements from different summands (remember that this is a direct sum) are linearly independent, and continues to be so after modding out by the maximal ideal. (you can't get any more relations after modding out, because we're considering the quotient as a $\mathfrak{o}_K/\pi_K\mathfrak{o}_K$-vector space, and any relation must come from the maximal ideal).
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2016-07-24 16:47:48
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https://zbmath.org/?q=an%3A1246.46001
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# zbMATH — the first resource for mathematics
Faber systems and their use in sampling, discrepancy, numerical integration. (English) Zbl 1246.46001
EMS Series of Lectures in Mathematics. Zürich: European Mathematical Society (EMS) (ISBN 978-3-03719-107-1/pbk). viii, 107 p. (2012).
In recent years the study of the efficiency of numerical approximation based on function evaluations (sampling) turned out to be important. The monograph [H. Triebel, Bases in function spaces, sampling, discrepancy, numerical integration. EMS Tracts in Mathematics 11 (2010; Zbl 1202.46002)] gives a good account of this problem. The present monograph continues the study from the one above with focus restricted to the use of Faber systems. The Faber system on $$(0,1)$$ consists of piecewise linear (hat) functions with respect to a dyadic partition. This function system has remarkable properties as being a (conditional) basis in the space of continuous functions, and becoming an unconditional basis in certain smoothness spaces of Sobolev type. Similar assertions hold true when turning to product Faber systems and appropriate function spaces. For the convenience of the reader the present monograph recalls the corresponding results from the above-mentioned monograph.
The objective in the present text can be summarized as follows. One feature of the Faber system is that only a finite number of point evaluations is required to compute a coefficient in the Faber system. The question arises whether all (natural candidate) Besov spaces have such a basis. The affirmative answer to this question would have impact on the sampling numbers in such spaces, say $$X$$, given by $g_{k}(\text{id}) := \inf\left\{\sup\left\{\| f- S_{k}(f) \|_{Y},\;f\in B_{X}\right\},\;\phi, \;x_{1},\dots,x_{k}\right\},$ where the supremum is taken over the unit ball $$B_{X}\subset X$$, and the infimum is taken over all choices of at most $$k$$ points in the domain $$\Omega$$, and reconstructions $$\phi$$.
A second motivation is the extension of previous results to weighted spaces on $$\mathbb R$$ or $$\mathbb R^{2}$$, respectively. Specifically weights of the form $$w^{\alpha}(x):= (1 + \left| x\right|^{2})^{\alpha/2}$$ are considered, or products thereof in the multivariate case. What can be said about the sampling numbers, numerical integration (with its integral numbers), or the discrepancy (with discrepancy numbers) in weighted spaces? The goal is not to extend to most general cases, but to clarify the influence of the weights on these questions. Finally, results for numerical discrepancy are given, clarifying some open problems in the affirmative.
To accomplish this goal the book starts with a detailed introduction in § 1 (Introduction, definitions, basic assertions), culminating in § 1.4 which discusses the objective of this study, as outlined before. In order to lay the foundations for the subsequent analysis, § 2 (Spaces on intervals) deals with the constructions and basic results for spaces on intervals. Here, the emphasis in on the discrepancy problem mentioned before. § 3 (Spaces on the real line) adapts the construction of the Faber system, originally given on $$(0,1)$$. Once this is accomplished one may ask similar questions as in the interval case, as in [loc. cit.], concerning the behavior of the sampling, integral and discrepancy numbers. The final § 4 (Spaces on the plane) extends the previous results to (anisotropic) weighted spaces. The basic results, Theorem 3.7 and Theorem 4.5, establish the asymptotics of the sampling numbers in weighted spaces. It is beyond the scope of this review to give a detailed account of this, but in loose terms the results asserts that for large (depending on smoothness and other parameters involved) weights $$\alpha$$ the situation in the weighted cases is as in the unweighted ones, whereas small weights $$\alpha$$ will have impact on the decay rate.
The monograph adds important information to problems of sampling, discrepancy and numerical integration in function spaces, and thus it contributes to a better understanding of sampling.
##### MSC:
46-02 Research exposition (monographs, survey articles) pertaining to functional analysis 46E35 Sobolev spaces and other spaces of “smooth” functions, embedding theorems, trace theorems 46B15 Summability and bases; functional analytic aspects of frames in Banach and Hilbert spaces 42C15 General harmonic expansions, frames 41A63 Multidimensional problems (should also be assigned at least one other classification number from Section 41-XX) 41A55 Approximate quadratures 41A30 Approximation by other special function classes 65D30 Numerical integration
Full Text:
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2021-06-13 17:44:44
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https://www.edumedia-sciences.com/en/media/757-squeeze-theorem-2
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# Squeeze theorem #2HTML5
## Summary
Limit at + ∞ or - ∞ of a sequence on a real interval, I, can be determined by comparison with two other functions whose limit is easily calculated.
This animation provides an illustration of the squeeze theorem applied to functions.
Click then slide the horizontal lines.
## Learning goals
• To know how to express and to apply the squeeze theorem for studying the limit of a sequence in infinity.
• To know how to express and to apply the comparison theorem for studying the limit of a sequence in infinity.
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2021-04-18 15:38:23
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http://ams.org/cgi-bin/bookstore/booksearch?fn=100&pg1=CN&s1=Silva_C_E&arg9=C._E._Silva
|
New Titles | FAQ | Keep Informed | Review Cart | Contact Us Quick Search (Advanced Search ) Browse by Subject General Interest Logic & Foundations Number Theory Algebra & Algebraic Geometry Discrete Math & Combinatorics Analysis Differential Equations Geometry & Topology Probability & Statistics Applications Mathematical Physics Math Education
Invitation to Ergodic Theory
C. E. Silva, Williams College, Williamstown, MA
SEARCH THIS BOOK:
Student Mathematical Library
2008; 262 pp; softcover
Volume: 42
ISBN-10: 0-8218-4420-2
ISBN-13: 978-0-8218-4420-5
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All Individuals: US\$36.80
Order Code: STML/42
Lectures on Fractal Geometry and Dynamical Systems - Yakov Pesin and Vaughn Climenhaga
Ergodic Theory, Groups, and Geometry - Robert J Zimmer and Dave Witte Morris
Mirror Symmetry and Algebraic Geometry - David A Cox and Sheldon Katz
This book is an introduction to basic concepts in ergodic theory such as recurrence, ergodicity, the ergodic theorem, mixing, and weak mixing. It does not assume knowledge of measure theory; all the results needed from measure theory are presented from scratch. In particular, the book includes a detailed construction of the Lebesgue measure on the real line and an introduction to measure spaces up to the Carathéodory extension theorem. It also develops the Lebesgue theory of integration, including the dominated convergence theorem and an introduction to the Lebesgue $$L^p$$spaces.
Several examples of a dynamical system are developed in detail to illustrate various dynamical concepts. These include in particular the baker's transformation, irrational rotations, the dyadic odometer, the Hajian-Kakutani transformation, the Gauss transformation, and the Chacón transformation. There is a detailed discussion of cutting and stacking transformations in ergodic theory. The book includes several exercises and some open questions to give the flavor of current research. The book also introduces some notions from topological dynamics, such as minimality, transitivity and symbolic spaces; and develops some metric topology, including the Baire category theorem.
Reviews
"The writing is crisp and clear. Proofs are written carefully with adequate levels of detail. Exercises are plentiful and well-integrated with the text."
-- MAA Reviews
"The author presents in a very pleasant and readable way an introduction to ergodic theory for measure-preserving transformations of probability spaces. In my opinion, the book provides guidelines, classical examples and useful ideas for an introductory course in ergodic theory to students that have not necessarily already been taught Lebesgue measure theory."
-- Elemente der Mathematik
"The book contains many (often easy or very easy) exercises, both in the text as well as at the end of each section."
-- Mathematical Reviews
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2014-07-22 22:06:01
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https://library.kiwix.org/bitcoin.stackexchange.com_en_all_2021-04/A/question/118.html
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## How much Bitcoin will I mine right now with hardware X?
113
48
Say I have a given piece of hardware (CPU/GPU). Where can I find out how much Bitcoin will I mine per day?
6
Bitcoin Mining Profitability Calculator: http://bitcoinX.com/profit
– kermit – 2012-02-23T08:45:20.933
60
Go to http://bitminter.com/test and click the "test start" button. If you have Java installed the miner should launch. Click "engine start" on your GPU(s) to start mining and the GUI will show how many bitcoins per day you will make (on average).
Note that you are actively mining in a pool without getting paid. This test page launches a version of the miner which is only meant to be used for a short time for testing.
Update for 2015: CPU and GPU mining are both long dead. When you buy ASIC mining hardware you will know its hashrate before you buy. Don't buy if you don't. Google "bitcoin mining calculator", input your hashrate, and it will tell you how much you can earn right now, on average. Note that the difficulty will change in the future so your earnings will not stay the same. Also note that with most pools actual earnings vary with luck. Finally, avoid the typical newbie mistake of confusing TH/s and GH/s. 1 TH/s = 1000 GH/s.
32018, GPU for btc is dead but not for other currency. – ewwink – 2018-01-31T05:20:00.697
31
There is a list here that is kept relatively up to date with the possible hashrate of various CPUs and GPUs
https://en.bitcoin.it/wiki/Mining_hardware_comparison
Once you know the number of hashes you can generate, you can use an online mining calculator and calculate the numbers of bitcoins you can mine per day (on average) and the current price of those bitcoins in other currencies.
http://tpbitcalc.appspot.com/
It is worth pointing out that when most people talk about "mining" bitcoins these days, they are talking about pooled mining. Depending on the fees of your pool you can make anywhere from 2%-10% less than expected. If you mine by yourself, the bitcoin you are expected to make has a high degree of variance.
https://en.bitcoin.it/wiki/Pooled_mining
These days when you say. do you mean 2018 or 2011? – YumYumYum – 2018-03-06T12:56:11.813
27
D = Difficulty
H = MHash/s
C = Reward (currently 12.5 BTC)
24 / (D * 2^32 / (H * 10^6) / 60 / 60) * C = BTC/day
@Naz No it does not. – Ken Sharp – 2021-01-27T22:24:51.350
Where does the 24 come from? – Ken Sharp – 2021-01-27T22:40:56.303
3Take note that the last figure (50) in this formula is the block reward, which at some point will halve to 25 and continue halving until it reaches zero, at which point mining will be subsidized only by transaction fees, which this formula also ignores. It's not unreasonable to ignore them for the time, since they're a very tiny fraction of mining income, but still :) – David Perry – 2011-10-12T20:46:15.513
Can you explain or reference what the 2^32 number is from and how its calculated? – nallenscott – 2016-05-22T13:31:11.047
A difficulty of D requires on average 2^32D hashes to solve. In fact, that's not entirely accurate. The exact number is 2^48/65535D. – Pieter Wuille – 2016-10-02T23:12:38.007
1@Pieter Wuille...So is his formula correct, because PEMDAS matters? I am trying to build my own calculator on my web page, for my own use, where ALL inputs are required. Many of these online calculators do not include, or allow input of variable items that are important. – dinotom – 2017-12-09T21:34:42.240
What is PEMDAS? – Pieter Wuille – 2017-12-09T21:43:07.253
Operation order; parentheses, exponents, multiply, divide, add, subtract – dinotom – 2017-12-09T23:27:31.927
@Pieter Wuille...PEMDAS is operation order; Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. I believe the correct formula is ...(DaysInMonth * Hashrate * BlockReward * SecsInDay) / (Difficulty * Exponent) ..to get the monthly mined coin rate – dinotom – 2017-12-14T11:50:16.200
The formula is correct, but one of the constants is slightly off (by about 0.0015%). – Pieter Wuille – 2017-12-14T18:03:40.953
this doesn't make sense .../60/60 means that the two 60 cancel out.... and so you end up with 24/(D2^32/(H10^6))*C – Naz – 2018-02-14T11:19:51.580
10
If:
D is the current difficulty
H is your hash rate in Mhash/s
B is block reward in BTC
Then you can expect to earn:
(H*B/D) * (60*60*24 * 65535 * 10^6 / 2^48)
= (H*B/D) * (5.662224e15 / 2^48) BTC per day (1)
or roughly:
(H*B/D) * 20.11626 BTC per day (2)
The current block reward B is a little over 12.5 BTC if you take transaction fees into account. (The transaction fees vary block by block; the core reward halves periodically, e.g. to 6.25 in mid-2020.)
(1) is exactly correct
(2) is an approximation, correct to 7 significant digits
This answer is slightly off due to assuming 65535 == 65536.
So what's the values of D, H, and B? There's a load of multiplications but none of them show 12.5 so it's impossible to disassemble this. – Ken Sharp – 2021-01-27T22:39:50.750
4
Yet another bitcoin mining profitability calculator - detailed and including a decline factor to compensate for mining hardware advancement (I am the author).
2
You should consult this wiki entry to know how many hashes per second can you generate with your hardware. Afterwards, just input the value into a bitcoin mining calculator and you will have an estimation of your payout ratio.
Please note that this calculator does not take difficulty into account so your earned value is likely to change over time. There are other calculators that try to predict the evolution of the difficulty, but they are not very accurate. Right now, difficulty depends on too many factors to be correctly predicted for a period longer than a couple of months.
1
I was confused by the profitability calculators I saw because I find it hard to get a feeling for how the difficulty influences my yield and also for where I can expect the difficulty to go, since it's such an abstract value.
So I now use what to me is more intuitive: there's a (more or less) fixed number of 3,600 BTC gained from mining each day, until the block reward halves in 2017. Those coins can be viewed as being shared by the whole network proportionally to the computing power of each player.
So, let's assume a network power of 360 Thash/s, which seems not far away as I write this, and makes for easy calculation, then for every Ghash/s that your equipment brings to the table, you get 0.01 BTC per day.
If you're solo mining, this is overly simplified, but the general direction of solo mining is that your actual yield will only on average match the theoretical one, so if you're much much smaller than the network (likely), then your variance gets so high that it becomes a lot like a lottery.
1
NiceHash has a nice calculator that will give you expected return on crypto mining, based on hardware and electricity price
-2
If you are interested in how your profitability will change as the network hashrate grows this calculator is the best IMO.
http://bitcoininsight.com/2013/04/05/bitcoin-mining-profitability-calculator/
Although you will have to make some guesses about how the network will grow, it should be more accurate than just a plain guess as to what the difficulty will be in X months time.
2This page 404's for me. Do you have a link that doesn't? – Nick ODell – 2013-04-15T00:06:54.183
1
I found the correct link and submitted an edit to this answer: http://bitcoininsight.com/2013/04/05/bitcoin-mining-profitability-calculator/
– Austin Burk – 2013-12-14T22:06:12.950
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2021-06-21 15:23:52
|
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|
https://math.stackexchange.com/questions/2214119/find-a-second-suitable-matrix-for-equation
|
Find a second suitable matrix for equation
Given the matrix:
$$\begin{pmatrix} 0 & 0 & 1 & 1\\ -2 & 2 & 3 & -3\\ 1 & -1 & -2 & 1 \\ \end{pmatrix}$$
1. Find an invertible matrix P and a Reduced row echelon form matrix D such that: $PA = D$
2. Find a bases for $ColA$
3. Find a second matrix $Q$ that: $P\neq Q$ and $QA = D$
I've managed to solve questions 1: $$D = \begin{pmatrix} 1 & -1 & 0 & 3\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0 \\ \end{pmatrix} , P = \begin{pmatrix} 0 & -2 & -3\\ 0 & -1 & -2\\ 1 & 1 & 2 \\ \end{pmatrix}$$
and question 2: $$\begin{bmatrix} 0 \\ 2 \\ 1 \\ \end{bmatrix}, \begin{bmatrix} 1 \\ 3 \\ -2 \\ \end{bmatrix}$$
But I don't know how to solve question 3.
$Q = A^{-1}D$
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2021-09-21 09:29:33
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https://www.techwhiff.com/issue/someone-help-if-u-know-the-answer-pls-put-the-step--451280
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# Someone Help if u know the answer pls put the step by step 6 = √v-2
1 answer
###### Question:
someone Help if u know the answer pls put the step by step
6 = √v-2
## Answers
1 answer
### How did the reformation impact the renaissance and what was the reformation. write a paragraph. plz help
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### Which of the following describes how an advertisement is produced? A А. An advertising company consults with the public before creating and distributing an ad. B An advertisement company creates an ad and then makes sure it is exposed to the maximum number of viewers. An advertising company collects information on the product and then provides the public with facts. D An advertising company checks the value of a product before creating an ad to support it.
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1 answer
### The price of Chive Corp. stock will be either $86 or$119 at the end of the year. Call options are available with one year to expiration. T-bills currently yield 5 percent. a. Suppose the current price of the company's stock is $97. What is the value of the call option if the exercise price is$85 per share? (Do not round intermediate calculations and round your answer to 2 decimal places, e.g., 32.16.) b. Suppose the exercise price is $115 and the current price of the company's stock is$97. Wh
The price of Chive Corp. stock will be either $86 or$119 at the end of the year. Call options are available with one year to expiration. T-bills currently yield 5 percent. a. Suppose the current price of the company's stock is $97. What is the value of the call option if the exercise price is$85 p...
1 answer
### PLZZZ HELP QUESTION:Find the slope between the 2 points (rise/run) What is the y-intercept Write the equation in y=mx+b
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### The Minnesota River is what is left from the River Warren. If you look at the land surrounding the Minnesota River, you will see it is U-shaped. From this, you can tell the River Warren was part of a glacier. part of mass wasting. a rivulet. an area of intense winds.
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### Can someone help me with this Question?
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### A right triangle is shown below. What is the value of x rounded to the nearest ten, if necessary?
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1 answer
### Write the following comparison as a ratio reduced to lowest terms. 5 nickels to 40 dimes
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1 answer
### I will gib brainliest if u help me >:C
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### Mark needs to edit many different parts of a presentation. Which of the following views does he need to be in to effectively edit the presentation? A. Slide Sorter B. Print Preview C. Slide Show D. Normal
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1 answer
### Plzzzzz helpppp meeeeee
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1 answer
### Just a lil science to make you smile
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1 answer
### Question Help 3.1.PS-10 The Stock Market Alexander is a stockbroker. He earns 14% commission each week. Last week he sold $7,100 worth of stocks. How much did he make last week in commission? If he averages that same amount each week, how much did he make in commission in 2011? Last week, he made$ in commission.
Question Help 3.1.PS-10 The Stock Market Alexander is a stockbroker. He earns 14% commission each week. Last week he sold $7,100 worth of stocks. How much did he make last week in commission? If he averages that same amount each week, how much did he make in commission in 2011? Last week, he made$ ...
2 answers
### What are the steps and structures involved in the process of cellular respiration?
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### Pls help it’s due today
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### Write 85% as a fraction in simplest form.
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### Which of the following is NOT a positive effect of modernization? a. longer life expectancies b. establishment of educational institutions c. higher rates of literacy d. weaker social relationships
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1 answer
### Help....................................................
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### Who has actually seen the Wizard of Oz in The Wonderful Wizard of Oz? A. Everyone in Oz . B. Only the Witches C. Only the Winkies UD. No one
Who has actually seen the Wizard of Oz in The Wonderful Wizard of Oz? A. Everyone in Oz . B. Only the Witches C. Only the Winkies UD. No one...
-- 0.017667--
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2022-08-08 08:19:10
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https://repository.uantwerpen.be/link/irua/127417
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Publication
Title
Precise estimates for the solution of stochastic functional differential equations with discontinuous initial data : part 1
Author
Abstract
In this work we have used the same introduction,notations and denitions as in [2]. Here we have proved a theorem in which we have established a uni- form error bound for the Euler approximation to the solution process of the Stochastic Funtional Dierential Equation (S.F.D.E.) (1.11) over the whole time interval [0; a]. This Theorem is an extension of the work of Kloeden and Platen ([6], Theorem 10.2.2) to S.F.D.E.'s with discontinuous initial data. We have calculated this uniform error bound by computing the dierence between the actual solution process and it's Euler approximation and we have found the upper bound for this dierence. We have also discussed the dependence of this dierence on the inital data.We have also proved that the Euler approx- imation of the solution process has the order of strong convergence = 0:5 see[6]chapters9and10.
Language
English
Source (journal)
International journal of innovative science, engineering & technology
Publication
2014
Volume/pages
1:8(2014), p. 179-191
Full text (open access)
UAntwerpen
Faculty/Department Research group Publication type Subject Affiliation Publications with a UAntwerp address
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2017-11-24 04:02:19
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https://hal.archives-ouvertes.fr/hal-01544442
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# Standard inelastic shocks and the dynamics of unilateral constraints
Abstract : This paper is devoted to mechanical systems with a finite number of degrees of freedom let $q^1, \ldots, q^n$ denote (possibly local) coordinates inthe configuration manifold $Q$. In addition to the constraints, bilateral and frictionless, which have permitted such a finite-dimensional parametrization of $Q$ , we assume the system submitted to a finite family of unilateral constraints whose geometrical effect is expressed by $v$ inequalities $f-\alpha(q) \leq 0$ defining a closed region $L$ of $Q$. As every greek index in the sequel, $α$ takes its values in the set $\{1,2,...,v\}$. The $v$ functions $f_α$ are supposed $C^1$, with nonzero gradients, at least in some neighborhood of the respective surfaces $f_α = 0$; for the sake of simplicity, we assume them independent of time.
Document type :
Book sections
Cited literature [27 references]
https://hal.archives-ouvertes.fr/hal-01544442
Contributor : Huong Le Thi <>
Submitted on : Monday, July 3, 2017 - 11:52:53 AM
Last modification on : Monday, August 26, 2019 - 11:10:03 AM
Long-term archiving on: : Friday, December 15, 2017 - 12:22:59 AM
### File
JJMoreau-Inelastic shocks.pdf
Publication funded by an institution
### Citation
Jean Jacques Moreau. Standard inelastic shocks and the dynamics of unilateral constraints. Unilateral Problems in Structural Analysis, 1983, 9783211818596. ⟨10.1007/978-3-7091-2632-5_9⟩. ⟨hal-01544442⟩
Record views
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2021-04-20 04:58:29
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https://www.msuperl.org/wikis/pcubed/doku.php?id=184_notes:examples:week6_charges_circuit&rev=1528744050
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184_notes:examples:week6_charges_circuit
This is an old revision of the document!
On the circuit shown below, draw how you would expect charge to distribute on the surface of the wire near the bends in the circuit.
Circuit with Bends
Facts
• Electric field is constant in the wire - created by the surface charges on the wire, not the battery.
• Electric field points along the direction of the wire, from positive to negative.
• Charges can build up on the surface of the wire.
Goal
• Draw charge distribution near the bends in the wire.
Representations
• We represent the circuit as in the example statement above.
• We will represent the charge distribution with little pluses ($+$) and minuses ($-$).
Assumption
In order to talk about individual charged particles in the circuit, we will have to make an assumption about the charged particles which are actually moving, the ones we think about when we think about “current”.
• The mobile charge carriers are positive.
It doesn't so much matter what we choose, but we want to be clear about the assumption, so that our discussion take place with this in mind, and so our reasoning will be consistent.
Note that we are only looking for the surface charges on the bends here. We need to figure out a way to distribute them so that the electric field stays constant in magnitude and along the direction of the wire. Let's consider the nature of the electric field near the positive end of the battery. In the wire, it will be pointing left into the bend. It will also have a large magnitude since it is so close to the battery. To reduce the magnitude, and ensure that the electric field points down after the bend, we place a lot of positive surface charges on the outer edge of the bend.
What would have happened if we did not place these charges here? Conventional current would follow the electric field. An electric field vector near the bend would point left, and instead of following the wire, positive charge would build up along the outer surface of the bend (the wider part of the bend) in the wire. Charge would continue to build up until we reach a point where an electric field vector near the bend points down! See below for a representation of how the surface charges near the bends in the wire would look. The electric field (and direction of conventional current) is shown with arrows.
Circuit with Surface Charges on the Bends
Notice that as you traverse the length of the wire, the surface charges become less positive and more negative, and eventually start to accumulate on the tighter part of the bend. We require each bend to be more negative than the previous one, so that the electric field on the straight parts of the wire points in the desired direction. And we start to draw the bends with negative charge on the tighter part, because all we need is for the wider part of the bend to be more positive than the tighter part. When drawing positive charge, it will be on the wider part. When drawing negative charge, it will be on the tighter part.
Note
The accumulation of charge around the bends might look slightly different if we had chosen for the mobile charge carriers to be negative in our assumption at the top of the solution. It's a worthwhile exercise to go through the reasoning again with the flipped assumption.
• 184_notes/examples/week6_charges_circuit.1528744050.txt.gz
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2021-09-22 08:31:10
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https://socratic.org/questions/how-is-water-represented-in-liquid-phase
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# How is water represented in liquid phase?
${H}_{2} O \left(l\right)$
$l$ stands for liquid, as $s$ stands for solid (ice in the case of water) and $g$ for gas (water vapour).
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2020-11-25 06:18:21
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https://thatsmaths.com/2023/02/
|
## Archive for February, 2023
### A Puzzle: Two-step Selection of a Digit
Here is a simple problem in probability.
(1) Pick a number k between 1 and 9. Assume all digits are equally likely.
(2) Pick a number m in the range from 1 to k.
What is the probability distribution for the number m?
A graph of the probability distribution is shown in the figure here.
Probability distribution for a decimal digit selected in a two-step process.
Can you derive a formula for this probability distribution?
Can you generalise it to the range from 1 to 10^n?
Can you relate this problem to Benford’s Law [described here]?
Solution, and more on Benford’s Law, next week.
### Weather Warnings in Glorious Technicolor
Severe weather affects us all and we need to know when to take action to protect ourselves and our property. We have become familiar with the colourful spectrum of warnings issued by Met Éireann.
For several years, Met Éireann has issued warnings of extreme weather. These depend on the severity of the meteorological event and the level of confidence in the forecast. They are formulated using forecasts produced by computer, algorithms that determine the likely impacts of extreme weather, and the expertise of the forecasters [TM242 or search for “thatsmaths” at irishtimes.com]. Continue reading ‘Weather Warnings in Glorious Technicolor’
### Ford Circles & Farey Series
Lester R Ford, Sr. (1886–1967).
American mathematician Lester Randolph Ford Sr. (1886–1967) was President of the Mathematical Association of America from 1947 to 1948 and editor of the American Mathematical Monthly during World War II. He is remembered today for the system of circles named in his honour.
For any rational number ${p/q}$ in reduced form (${p}$ and ${q}$ coprime), a Ford circle is a circle with center at ${(p/q,1/(2q^{2}))}$ and radius ${1/(2q^{2})}$. There is a Ford circle associated with every rational number. Every Ford circle is tangent to the horizontal axis and each two Ford circles are either tangent or disjoint from each other.
### From Wave Equations to Modern Telecoms
Mathematics has an amazing capacity to help us to understand the physical world. Just consider the profound implications of Einstein’s simple equation ${E = m c^2}$. Another example is the wave equation derived by Scottish mathematical physicist James Clerk Maxwell. Our modern world would not exist without the knowledge encapsulated in Maxwell’s equations. Continue reading ‘From Wave Equations to Modern Telecoms’
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2023-03-30 05:44:16
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http://web2.0calc.com/questions/find-a-base-7-three-digit-number-which-has-its-digits-reversed-when-expressed-in-base-9-you-do-not-need-to-indicate-the-base-with-a-subscr
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+0
# Find a base 7 three-digit number which has its digits reversed when expressed in base 9. (You do not need to indicate the base with a subscr
0
757
2
+1766
Find a base 7 three-digit number which has its digits reversed when expressed in base 9. (You do not need to indicate the base with a subscript for this answer.)
Mellie Jun 7, 2015
#2
+91469
+10
Let the number be XYZ XY and Z are all 1 digit numbers between 0 and 6 inclusive
X and Z cannot by 0
Z must be smaller than X
So Y could possible be 0,12,3,4,5,6
So X could possible be 2,3,4,5,6
So Z could possible be 1,2,3,4,5
$$\\X*7^2+Y*7+Z=Z*9^2+Y*9+X\\\\ 49X+7Y+Z=81Z+9Y+X\\\\$$
I might just use trial and error Mellie
Try Z=1
49X+7Y+1=81*1+9Y+X
48X-2Y=80
24X-Y= 40 Multiples of 24 are 24,48, Neither of these - an allowed y =40
so Z is not 1
Try Z=2
49X+7Y+2=81*2+9Y+X
48X-2Y=160
24X-Y=80 Multiples of 24 are 24,48,72,96 Neither of these - an allowed y =80
so Z is not 2
Try Z=3
49X+7Y+3=81*3+9Y+X
48X-2Y=240
24X-Y=120 The first multiple of 24 that is BIGGER than or equal to 120 is 120
SO Z can be 3, X=5 and Y=0
$$\\\mathbf{503_7=305_9}$$
check
$${\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{49}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}} = {\mathtt{248}}$$
$${\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{81}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}} = {\mathtt{248}}$$ (Edited: Thanks Chris )
There might be a much quicker way of doing this
Melody Jun 8, 2015
Sort:
#2
+91469
+10
Let the number be XYZ XY and Z are all 1 digit numbers between 0 and 6 inclusive
X and Z cannot by 0
Z must be smaller than X
So Y could possible be 0,12,3,4,5,6
So X could possible be 2,3,4,5,6
So Z could possible be 1,2,3,4,5
$$\\X*7^2+Y*7+Z=Z*9^2+Y*9+X\\\\ 49X+7Y+Z=81Z+9Y+X\\\\$$
I might just use trial and error Mellie
Try Z=1
49X+7Y+1=81*1+9Y+X
48X-2Y=80
24X-Y= 40 Multiples of 24 are 24,48, Neither of these - an allowed y =40
so Z is not 1
Try Z=2
49X+7Y+2=81*2+9Y+X
48X-2Y=160
24X-Y=80 Multiples of 24 are 24,48,72,96 Neither of these - an allowed y =80
so Z is not 2
Try Z=3
49X+7Y+3=81*3+9Y+X
48X-2Y=240
24X-Y=120 The first multiple of 24 that is BIGGER than or equal to 120 is 120
SO Z can be 3, X=5 and Y=0
$$\\\mathbf{503_7=305_9}$$
check
$${\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{49}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}} = {\mathtt{248}}$$
$${\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{81}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}} = {\mathtt{248}}$$ (Edited: Thanks Chris )
There might be a much quicker way of doing this
Melody Jun 8, 2015
#3
+81031
+5
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2018-01-21 08:48:44
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https://davefoord.wordpress.com/category/how-do-i/
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• ## Email Subscription
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• ## del.icio.us
In November 2011, I wrote a blog post titled ‘How to automatically pull data between different Google Spreadsheets‘ – which was based on a feature called ‘ImportRange’. Although nearly 5 years old, this particular blog post is one of my most frequented and certainly the most commented on post that I have ever written.
I have recently been introduced to a new add on feature called ‘Import Sheet’ – which can be found at https://importsheet.com/ – and this does exactly what it says on the tin. It allows you to easily import (or export) a sheet from one file to another. So for example, I use google sheets to log the work that I do for each of my clients, with each client having a separate workbook (this means that I can share that workbook with them, without them seeing other contracts that I work on). I can then have a master dashboard, which imports copies of these sheets into a single location, and using simple functions such as the = function, I can then pull key data out of each of the sheets into my dashboard. Using Importsheet rather than the ImportRange feature is much easier, quicker and less likely to have problems.
The add on is a commercial tool, that does have a pricing plan, however the free version does what most people will want. Obviously I have no idea how the pricing plan may change in the future – they may choose to get rid, or reduce the functionality of the free version, so I wouldn’t suggest that people invest lots of time creating high stakes activity with this add on (unless they are prepared to pay in the future) – but certainly for the moment, this looks rather neat.
## Creating equations in online environments
I am currently working on a project, where I am creating web based resources (using Moodle) to teach maths. As part of this process, I need to create properly laid out formulas (or formulae if you prefer the alternate acceptable plural of formula). We are using what is becoming a widely accepted standard of MathJax which in turn supports the use of something called LaTex to create the desired formulas. So for example if I wanted to create something that looked like:
$y = 3x^{2}+5x+\frac{2}{3}$
I would enter this into the editor using the code:
$$y = 3x^{2}+5x+\frac{2}{3}$$
or
$y = 3x^{2}+5x+\frac{2}{3}$
At first I started to learn the exact syntax and would translate what I wanted into either of the codes above. This proved to be both time consuming and prone to mistakes. Then I discovered an excellent website that helps me do this:
https://www.codecogs.com/latex/eqneditor.php
This website gives me a box into which I construct the equation that I want – above the box is a huge suite of grey buttons which each represent a different mathematical function or options. These take a bit of time to learn, but quite quickly one gets the hang of this, and using the buttons and adding the numbers / letters that you require you can quite quickly create the desired formula that you want. Underneath the white box, your formula is displayed as it will appear, so it is possible to see what you are doing, and check that this is correct.
Once you are happy with what you have created, at the bottom of the screen (in a cream coloured box) is the option to choose the export style that you want – so if you need LaTex, you choose that, if embedding into WordPress, you choose WordPress etc. You then copy the code beneath this, and paste into the editor of whatever you are using.
One of the facilities within the editor, is to create correctly aligned equations:
Which in LaTex is created with the code:
\begin{align*} x+3 &= 7\\ x &= 7-3\\ x &= 4 \end{align*}
This is very hard to manually write out, but quite easy using the codecogs website. The button for this, is the bottom right button on toolbar (letters n and r in brackets) – then under that is a button that has “y=…” as the text, and when you hover over it, it tells you that it is the align tool.
For anyone who is using mathematical formulas regularly this is a really neat tool.
## Creating a YouTube based discussion activity in Moodle
I run a lot of training on effective uses of a VLE (usually Moodle) and one of the easiest activities that I show, is finding a video on YouTube, and then embedding this into a forum activity within the VLE.
The reasons for doing this are:
1. By embedding the video (rather than simply linking to it) – we remove all the distractions, adverts, etc. that appear on YouTube around the edges.
2. By adding this as a discussion activity, we ask the students a question – this will focus their attention whilst watching the video, rather than just passively ‘absorbing’ it.
It doesn’t matter if students don’t actually post their answers to the forum (although useful if they do), as they will still benefit from watching the video with the question in their mind.
The following video goes through the steps of how to embed the video, and the basic settings within a Moodle forum activity.
And if you want to only show a portion of the video you can always identify the exact start and end points that you want to play, by following these instructions:
## Trimming and Embedding a YouTube Video into PowerPoint
I have blogged many times in the past about things to do with PowerPoint, including how to embed a YouTube video, or how to use TubeChop to embed a YouTube video.
In more recent versions of PowerPoint (2013 and 2016), the ability to embed a YouTube video has been made easier, and the following video will take you through the steps:
Although easier to do than in the past, this technique has been unreliable for some people in some organisations, so I always recommend to people to paste the video’s URL onto the slide somewhere as a live link, so if this doesn’t work, you have the fall back of simply accessing the video via the YouTube website.
This technique is showing how to embed the video – this means you still need access to the Internet when viewing the presentation, and it won’t work if the organisation blocks YouTube.
This technique replaces the older method of using the shockwave flash object, or using TubeChop to trim the video.
The Moodle Glossary is one of the simplest activities to use within Moodle. It’s primary purpose is to set up a glossary that the students populate or at least add to, rather than the tutor populating it – however there are occasions when it is useful for the tutor to populate the glossary. This could be if the tutor wants to provide the students with a ‘correct’ list of technical terms and their definitions, or if the tutor wants to create a crossword using the Moodle Game Activity plugin – which pulls the data out of a glossary activity.
If the tutor is populating the glossary, they could enter the data manually item by item, but this is very time consuming, especially if they already have the data to hand in a spreadsheet or similar. Luckily there is a way to bulk upload these items. It is a little complex, but once you have done a few not too bad, and certainly a lot easier than manually typing in lots of items.
Firstly – we need to create the glossary, the following video from Moodle, goes through this step:
Then the clever part is importing the list of terms from an external source. This is covered in this video:
The XML converter that is used, can be located here:
https://moodle.org/mod/forum/discuss.php?d=91224#p489666 with the direct link to the actual file itself being:
https://moodle.org/pluginfile.php/159/mod_forum/attachment/489666/glossaryXMLconverter_html4.zip
You only need to download and unzip this once – as long as you can remember where you have saved it to. Although it may seem a little convoluted at first, this technique will save serious amounts of time compared to manually entering lots of data.
The only other thing to note, is the default settings of the glossary may allow the students to add their own items to this glossary, if you don’t want this to happen, you can stop this by using either of the following options:
1. In the glossary settings – make it so the ‘Approved by default’ option is set to ‘no’.
2. Go into the permissions for that glossary and next to where is says ‘Create new entries’ delete students from the list of roles that can do this.
## Correcting problem with Outlook 2013 not always sending messages
I have recently upgraded my windows computer from Windows 8 to Windows 10, which in general is great, but it did cause a problem with outlook 2013, in that some messages wouldn’t send. The problem wasn’t consistent, which was confusing, I tried searching the internet for help, but the solutions offered all seem very complicated, so I contacted my web host, who talked me through the solution, which worked, so I thought that I would share this here.
Please note – I am not technically orientated, I am just repeating the steps for my benefit in case I need it in the future – if you follow these steps, you do so at your own risk, and if it doesn’t solve your problem, please don’t ask me – I won’t know the answer.
So to solve the problem:
1. I right clicked on Windows icon.
2. I selected command prompt (admin)
3. I typed “sfc/scannow” then enter.
4. I let this run the scan (took about 15 minutes).
5. Once scan had finished I closed down Outlook and then restarted it.
As mentioned earlier, I am not technically minded, so I have no idea what is happening with the above apart from it appeared to fix my problem.
## How to trim a YouTube video and embed it into WordPress
Regular followers of my blog, will know that I have previously blogged about different ways of trimming or cropping YouTube videos to use in different situations, e.g:
Trimming videos can be really valuable, as often (in education) there may be a key message or element in a video, that we want to draw attention to, without having to show the entire video, so carefully selecting sections of the video can drastically improve the impact of using that video as a resource.
Up until recently, something that I couldn’t do was to get this to work in WordPress, I could only embed the whole video – however today I found out that I can do trim a video, if you follow these instructions exactly:
1. Locate the video you want on YouTube.
2. Under the video choose ‘Share’ and then ‘Embed’.
3. Underneath the preview, there will be some tick boxes – make sure the one called ‘Show suggested videos when the video finishes‘ is unticked (this is really important).
4. Copy the embed code that is above the preview.
5. Go into your WordPress post, and into the HTML editor.
6. Paste the copied embed code in the correct position.
8. Go back to edit your post, and again into the HTML editor.
9. The code that you pasted in, will have been changed, towards the end of it, locate the text rel=0.
10. Immediately after rel=0 add &start=xxx&end=yyy (where xxx is the number of seconds you want the video to start at and yyy is the number of seconds you want the video to end at).
11. Preview your post – if it works then publish.
So – if I have a video that I want to start at 2:35 and end at 3:15 – I convert these into seconds (2:35 = 155, 3:15 = 195) – and the end of the code will change from:
….?rel=0]
to
….?rel=0&start=155&end=195]
Below is an example of one of my YouTube videos, trimmed to start at 155 seconds and end at 195 seconds.
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2017-02-25 06:46:09
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https://socratic.org/questions/how-to-construct-triangle-lmn-such-that-ln-8-cm-and-lmn-80-degrees-and-lm-mn-3-c
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# How to construct triangle LMN , such that LN = 8 cm and ∠ LMN = 80 degrees and LM- MN= 3 cm?
Feb 13, 2016
Use $L M = 7.461 c m$ and $M N = 4.461 c m$
#### Explanation:
Calling LM as $x$ and MN as $y$
$x - y = 3$ => $y = x - 3$
As $\angle L M N$ is the angle between x and y
Applying the Law of Cosines:
${8}^{2} = {x}^{2} + {y}^{2} - 2 x \cdot y \cdot \cos {80}^{\circ}$
$64 = {x}^{2} + {\left(x - 3\right)}^{2} - 2 x \left(x - 3\right) \cos {80}^{\circ}$
${x}^{2} + {x}^{2} - 6 x + 9 - 2 {x}^{2} \cos {80}^{\circ} + 6 x \cos {80}^{\circ} - 64 = 0$
$\left(2 - 2 \cos {80}^{\circ}\right) {x}^{2} + \left(- 6 + 6 \cos {80}^{\circ}\right) x - 55 = 0$
$\Delta = 36 + 36 {\cos}^{2} {80}^{\circ} - 72 \cos {80}^{\circ} + 440 - 440 \cos {80}^{\circ}$
$\Delta = 476 + 36 {\cos}^{2} {80}^{\circ} - 512 \cos {80}^{\circ}$
$\Delta \cong 388.178$ => $\sqrt{\Delta} \cong 19.702$
x=(6(1-cos 80^@)+-19.702)/(4(1-cos 80^@)
$\to {x}_{1} = 7.461 c m$ => $y = 7.461 - 3$ => $y = 4.461 c m$
$\to {x}_{2} = - 4.461$ (not valid)
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2022-12-07 07:12:40
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http://mathhelpforum.com/calculus/78216-derivative-x-raised-function.html
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# Thread: Derivative of x raised to a function
1. ## Derivative of x raised to a function
Differentiate:
$\displaystyle y=x^ {2-X}$
2. Originally Posted by ahhh
Differentiate:
$\displaystyle y=x^ {2-X}$
Here's a start (assuming you're familiar with implicit differentiation):
$\displaystyle y = x^{2 - x} \Rightarrow \ln y = (2 - x) \ln (x) \Rightarrow \frac{1}{y} \frac{dy}{dx} = - \ln (x) + \frac{2-x}{x}$.
3. $\displaystyle y = x^{2 - x} \Rightarrow \ln y = (2 - x) \ln (x) \Rightarrow \frac{1}{y} \frac{dy}{dx} = - \ln (x) + \frac{2-x}{x}$
Ohhhh...how can you tell when you need to use logarithmic differentation?
I understand the process now. Thanks!!
4. Originally Posted by ahhh
$\displaystyle y = x^{2 - x} \Rightarrow \ln y = (2 - x) \ln (x) \Rightarrow \frac{1}{y} \frac{dy}{dx} = - \ln (x) + \frac{2-x}{x}$$\displaystyle$
Ohhhh...how can you tell when you need to use logarithmic differentation?
I understand the process now. Thanks!!
$\displaystyle y = f(x)^{g(x)}$ is always a typical candidate.
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2018-06-19 09:02:03
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|
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=130&t=41209
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$\Delta U=q+w$
Claudeth Martinez 1D
Posts: 52
Joined: Fri Sep 28, 2018 12:18 am
Alma Flores 1D
Posts: 64
Joined: Wed Nov 08, 2017 3:01 am
Adiabatic means that it is not permitting or accompanied by the passage of energy as heat.
sallina_yehdego 2E
Posts: 75
Joined: Sun Apr 29, 2018 3:00 am
Adiabatic means q=0, which means delta U=w
1K Kevin
Posts: 47
Joined: Fri Sep 28, 2018 12:24 am
Adiabatic is a quick process that does not allows for heat transfer, but the temperature pressure and volume can all change. All change in energy would be from Work.
Hilda Sauceda 3C
Posts: 76
Joined: Fri Sep 28, 2018 12:24 am
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2021-03-07 18:52:50
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https://docs.python.org/fr/3/library/shutil.html
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# shutil --- Opérations de haut niveau sur les fichiers¶
Code source : Lib/shutil.py
Le module shutil propose des opérations de haut niveau sur les fichiers et ensembles de fichiers. En particulier, des fonctions pour copier et déplacer les fichiers sont proposées. Pour les opérations individuelles sur les fichiers, reportez-vous au module os.
Avertissement
Même les fonctions de copie haut niveau (shutil.copy(), shutil.copy2()) ne peuvent copier toutes les métadonnées des fichiers.
Sur les plateformes POSIX, cela signifie que le propriétaire et le groupe du fichier sont perdus, ainsi que les ACLs. Sur Mac OS, le clonage de ressource et autres métadonnées ne sont pas utilisés. Cela signifie que les ressources seront perdues et que le type de fichier et les codes créateur ne seront pas corrects. Sur Windows, les propriétaires des fichiers, ACLs et flux de données alternatifs ne sont pas copiés.
## Opérations sur les répertoires et les fichiers¶
shutil.copyfileobj(fsrc, fdst[, length])
Copie le contenu de l'objet fichier fsrc dans l'objet fichier fdst. L'entier length, si spécifié, est la taille du tampon. En particulier, une valeur de length négative signifie la copie des données sans découper la source en morceaux ; par défaut les données sont lues par morceaux pour éviter la consommation mémoire non-contrôlée. À noter que si la position courante dans l'objet fsrc n'est pas 0, seul le contenu depuis la position courante jusqu'à la fin est copié.
shutil.copyfile(src, dst, *, follow_symlinks=True)
Copy the contents (no metadata) of the file named src to a file named dst and return dst in the most efficient way possible. src and dst are path-like objects or path names given as strings.
dst must be the complete target file name; look at copy() for a copy that accepts a target directory path. If src and dst specify the same file, SameFileError is raised.
La cible doit être accessible en écriture, sinon l'exception OSError est levée. Si dst existe déjà, il est remplacé. Les fichiers spéciaux comme les périphériques caractères ou bloc ainsi que les tubes (pipes) ne peuvent pas être copiés avec cette fonction.
Si follow_symlinks est faux et src est un lien symbolique, un nouveau lien symbolique est créé au lieu de copier le fichier pointé par src.
Raises an auditing event shutil.copyfile with arguments src, dst.
Modifié dans la version 3.3: IOError était levée au lieu de OSError. Ajout de l'argument follow_symlinks. Maintenant renvoie dst.
Modifié dans la version 3.4: Lève SameFileError au lieu de Error. Puisque la première est une sous-classe de la seconde, le changement assure la rétrocompatibilité.
Modifié dans la version 3.8: Platform-specific fast-copy syscalls may be used internally in order to copy the file more efficiently. See Platform-dependent efficient copy operations section.
exception shutil.SameFileError
Cette exception est levée si la source et la destination dans copyfile() sont le même fichier.
Nouveau dans la version 3.4.
shutil.copymode(src, dst, *, follow_symlinks=True)
Copy the permission bits from src to dst. The file contents, owner, and group are unaffected. src and dst are path-like objects or path names given as strings. If follow_symlinks is false, and both src and dst are symbolic links, copymode() will attempt to modify the mode of dst itself (rather than the file it points to). This functionality is not available on every platform; please see copystat() for more information. If copymode() cannot modify symbolic links on the local platform, and it is asked to do so, it will do nothing and return.
Raises an auditing event shutil.copymode with arguments src, dst.
Modifié dans la version 3.3: L'argument follow_symlinks a été ajouté.
shutil.copystat(src, dst, *, follow_symlinks=True)
Copy the permission bits, last access time, last modification time, and flags from src to dst. On Linux, copystat() also copies the "extended attributes" where possible. The file contents, owner, and group are unaffected. src and dst are path-like objects or path names given as strings.
Si follow_symlinks est faux et src et dst représentent des liens symboliques, copystat() agit sur les liens symboliques au lieu des fichiers cibles — elle lit les informations du lien symbolique de src et les écrit vers la destination pointée par dst.
Note
Toutes les plateformes n'offrent pas la possibilité d'examiner et modifier les liens symboliques. Python peut vous informer des fonctionnalités effectivement disponibles.
• Si os.chmod in os.supports_follow_symlinks est True, copystat() peut modifier les octets de droits d'accès du lien symbolique.
• If os.utime in os.supports_follow_symlinks is True, copystat() can modify the last access and modification times of a symbolic link.
• If os.chflags in os.supports_follow_symlinks is True, copystat() can modify the flags of a symbolic link. (os.chflags is not available on all platforms.)
On platforms where some or all of this functionality is unavailable, when asked to modify a symbolic link, copystat() will copy everything it can. copystat() never returns failure.
Please see os.supports_follow_symlinks for more information.
Raises an auditing event shutil.copystat with arguments src, dst.
Modifié dans la version 3.3: Added follow_symlinks argument and support for Linux extended attributes.
shutil.copy(src, dst, *, follow_symlinks=True)
Copies the file src to the file or directory dst. src and dst should be path-like objects or strings. If dst specifies a directory, the file will be copied into dst using the base filename from src. Returns the path to the newly created file.
If follow_symlinks is false, and src is a symbolic link, dst will be created as a symbolic link. If follow_symlinks is true and src is a symbolic link, dst will be a copy of the file src refers to.
copy() copies the file data and the file's permission mode (see os.chmod()). Other metadata, like the file's creation and modification times, is not preserved. To preserve all file metadata from the original, use copy2() instead.
Raises an auditing event shutil.copyfile with arguments src, dst.
Raises an auditing event shutil.copymode with arguments src, dst.
Modifié dans la version 3.3: Added follow_symlinks argument. Now returns path to the newly created file.
Modifié dans la version 3.8: Platform-specific fast-copy syscalls may be used internally in order to copy the file more efficiently. See Platform-dependent efficient copy operations section.
shutil.copy2(src, dst, *, follow_symlinks=True)
Identical to copy() except that copy2() also attempts to preserve file metadata.
When follow_symlinks is false, and src is a symbolic link, copy2() attempts to copy all metadata from the src symbolic link to the newly-created dst symbolic link. However, this functionality is not available on all platforms. On platforms where some or all of this functionality is unavailable, copy2() will preserve all the metadata it can; copy2() never raises an exception because it cannot preserve file metadata.
copy2() uses copystat() to copy the file metadata. Please see copystat() for more information about platform support for modifying symbolic link metadata.
Raises an auditing event shutil.copyfile with arguments src, dst.
Raises an auditing event shutil.copystat with arguments src, dst.
Modifié dans la version 3.3: Added follow_symlinks argument, try to copy extended file system attributes too (currently Linux only). Now returns path to the newly created file.
Modifié dans la version 3.8: Platform-specific fast-copy syscalls may be used internally in order to copy the file more efficiently. See Platform-dependent efficient copy operations section.
shutil.ignore_patterns(*patterns)
This factory function creates a function that can be used as a callable for copytree()'s ignore argument, ignoring files and directories that match one of the glob-style patterns provided. See the example below.
shutil.copytree(src, dst, symlinks=False, ignore=None, copy_function=copy2, ignore_dangling_symlinks=False, dirs_exist_ok=False)
Recursively copy an entire directory tree rooted at src to a directory named dst and return the destination directory. dirs_exist_ok dictates whether to raise an exception in case dst or any missing parent directory already exists.
Permissions and times of directories are copied with copystat(), individual files are copied using copy2().
If symlinks is true, symbolic links in the source tree are represented as symbolic links in the new tree and the metadata of the original links will be copied as far as the platform allows; if false or omitted, the contents and metadata of the linked files are copied to the new tree.
When symlinks is false, if the file pointed by the symlink doesn't exist, an exception will be added in the list of errors raised in an Error exception at the end of the copy process. You can set the optional ignore_dangling_symlinks flag to true if you want to silence this exception. Notice that this option has no effect on platforms that don't support os.symlink().
If ignore is given, it must be a callable that will receive as its arguments the directory being visited by copytree(), and a list of its contents, as returned by os.listdir(). Since copytree() is called recursively, the ignore callable will be called once for each directory that is copied. The callable must return a sequence of directory and file names relative to the current directory (i.e. a subset of the items in its second argument); these names will then be ignored in the copy process. ignore_patterns() can be used to create such a callable that ignores names based on glob-style patterns.
If exception(s) occur, an Error is raised with a list of reasons.
If copy_function is given, it must be a callable that will be used to copy each file. It will be called with the source path and the destination path as arguments. By default, copy2() is used, but any function that supports the same signature (like copy()) can be used.
Raises an auditing event shutil.copytree with arguments src, dst.
Modifié dans la version 3.3: Copy metadata when symlinks is false. Now returns dst.
Modifié dans la version 3.2: Added the copy_function argument to be able to provide a custom copy function. Added the ignore_dangling_symlinks argument to silent dangling symlinks errors when symlinks is false.
Modifié dans la version 3.8: Platform-specific fast-copy syscalls may be used internally in order to copy the file more efficiently. See Platform-dependent efficient copy operations section.
Nouveau dans la version 3.8: The dirs_exist_ok parameter.
shutil.rmtree(path, ignore_errors=False, onerror=None)
Delete an entire directory tree; path must point to a directory (but not a symbolic link to a directory). If ignore_errors is true, errors resulting from failed removals will be ignored; if false or omitted, such errors are handled by calling a handler specified by onerror or, if that is omitted, they raise an exception.
Note
On platforms that support the necessary fd-based functions a symlink attack resistant version of rmtree() is used by default. On other platforms, the rmtree() implementation is susceptible to a symlink attack: given proper timing and circumstances, attackers can manipulate symlinks on the filesystem to delete files they wouldn't be able to access otherwise. Applications can use the rmtree.avoids_symlink_attacks function attribute to determine which case applies.
If onerror is provided, it must be a callable that accepts three parameters: function, path, and excinfo.
The first parameter, function, is the function which raised the exception; it depends on the platform and implementation. The second parameter, path, will be the path name passed to function. The third parameter, excinfo, will be the exception information returned by sys.exc_info(). Exceptions raised by onerror will not be caught.
Raises an auditing event shutil.rmtree with argument path.
Modifié dans la version 3.3: Added a symlink attack resistant version that is used automatically if platform supports fd-based functions.
Modifié dans la version 3.8: On Windows, will no longer delete the contents of a directory junction before removing the junction.
Indicates whether the current platform and implementation provides a symlink attack resistant version of rmtree(). Currently this is only true for platforms supporting fd-based directory access functions.
Nouveau dans la version 3.3.
shutil.move(src, dst, copy_function=copy2)
Recursively move a file or directory (src) to another location (dst) and return the destination.
If the destination is an existing directory, then src is moved inside that directory. If the destination already exists but is not a directory, it may be overwritten depending on os.rename() semantics.
If the destination is on the current filesystem, then os.rename() is used. Otherwise, src is copied to dst using copy_function and then removed. In case of symlinks, a new symlink pointing to the target of src will be created in or as dst and src will be removed.
If copy_function is given, it must be a callable that takes two arguments src and dst, and will be used to copy src to dst if os.rename() cannot be used. If the source is a directory, copytree() is called, passing it the copy_function(). The default copy_function is copy2(). Using copy() as the copy_function allows the move to succeed when it is not possible to also copy the metadata, at the expense of not copying any of the metadata.
Raises an auditing event shutil.move with arguments src, dst.
Modifié dans la version 3.3: Added explicit symlink handling for foreign filesystems, thus adapting it to the behavior of GNU's mv. Now returns dst.
Modifié dans la version 3.5: Added the copy_function keyword argument.
Modifié dans la version 3.8: Platform-specific fast-copy syscalls may be used internally in order to copy the file more efficiently. See Platform-dependent efficient copy operations section.
Modifié dans la version 3.9: Accepts a path-like object for both src and dst.
shutil.disk_usage(path)
Return disk usage statistics about the given path as a named tuple with the attributes total, used and free, which are the amount of total, used and free space, in bytes. path may be a file or a directory.
Nouveau dans la version 3.3.
Modifié dans la version 3.8: On Windows, path can now be a file or directory.
Disponibilité : Unix, Windows.
shutil.chown(path, user=None, group=None)
Change owner user and/or group of the given path.
user can be a system user name or a uid; the same applies to group. At least one argument is required.
See also os.chown(), the underlying function.
Raises an auditing event shutil.chown with arguments path, user, group.
Disponibilité : Unix.
Nouveau dans la version 3.3.
shutil.which(cmd, mode=os.F_OK | os.X_OK, path=None)
Return the path to an executable which would be run if the given cmd was called. If no cmd would be called, return None.
mode is a permission mask passed to os.access(), by default determining if the file exists and executable.
When no path is specified, the results of os.environ() are used, returning either the "PATH" value or a fallback of os.defpath.
On Windows, the current directory is always prepended to the path whether or not you use the default or provide your own, which is the behavior the command shell uses when finding executables. Additionally, when finding the cmd in the path, the PATHEXT environment variable is checked. For example, if you call shutil.which("python"), which() will search PATHEXT to know that it should look for python.exe within the path directories. For example, on Windows:
>>> shutil.which("python")
'C:\\Python33\\python.EXE'
Nouveau dans la version 3.3.
Modifié dans la version 3.8: The bytes type is now accepted. If cmd type is bytes, the result type is also bytes.
exception shutil.Error
This exception collects exceptions that are raised during a multi-file operation. For copytree(), the exception argument is a list of 3-tuples (srcname, dstname, exception).
### Platform-dependent efficient copy operations¶
Starting from Python 3.8, all functions involving a file copy (copyfile(), copy(), copy2(), copytree(), and move()) may use platform-specific "fast-copy" syscalls in order to copy the file more efficiently (see bpo-33671). "fast-copy" means that the copying operation occurs within the kernel, avoiding the use of userspace buffers in Python as in "outfd.write(infd.read())".
On macOS fcopyfile is used to copy the file content (not metadata).
On Linux os.sendfile() is used.
On Windows shutil.copyfile() uses a bigger default buffer size (1 MiB instead of 64 KiB) and a memoryview()-based variant of shutil.copyfileobj() is used.
If the fast-copy operation fails and no data was written in the destination file then shutil will silently fallback on using less efficient copyfileobj() function internally.
Modifié dans la version 3.8.
### copytree example¶
This example is the implementation of the copytree() function, described above, with the docstring omitted. It demonstrates many of the other functions provided by this module.
def copytree(src, dst, symlinks=False):
names = os.listdir(src)
os.makedirs(dst)
errors = []
for name in names:
srcname = os.path.join(src, name)
dstname = os.path.join(dst, name)
try:
elif os.path.isdir(srcname):
else:
copy2(srcname, dstname)
# XXX What about devices, sockets etc.?
except OSError as why:
errors.append((srcname, dstname, str(why)))
# catch the Error from the recursive copytree so that we can
# continue with other files
except Error as err:
errors.extend(err.args[0])
try:
copystat(src, dst)
except OSError as why:
# can't copy file access times on Windows
if why.winerror is None:
errors.extend((src, dst, str(why)))
if errors:
raise Error(errors)
Another example that uses the ignore_patterns() helper:
from shutil import copytree, ignore_patterns
copytree(source, destination, ignore=ignore_patterns('*.pyc', 'tmp*'))
This will copy everything except .pyc files and files or directories whose name starts with tmp.
Another example that uses the ignore argument to add a logging call:
from shutil import copytree
import logging
def _logpath(path, names):
logging.info('Working in %s', path)
return [] # nothing will be ignored
copytree(source, destination, ignore=_logpath)
### rmtree example¶
This example shows how to remove a directory tree on Windows where some of the files have their read-only bit set. It uses the onerror callback to clear the readonly bit and reattempt the remove. Any subsequent failure will propagate.
import os, stat
import shutil
"Clear the readonly bit and reattempt the removal"
os.chmod(path, stat.S_IWRITE)
func(path)
## Archiving operations¶
Nouveau dans la version 3.2.
Modifié dans la version 3.5: Added support for the xztar format.
High-level utilities to create and read compressed and archived files are also provided. They rely on the zipfile and tarfile modules.
shutil.make_archive(base_name, format[, root_dir[, base_dir[, verbose[, dry_run[, owner[, group[, logger]]]]]]])
Create an archive file (such as zip or tar) and return its name.
base_name is the name of the file to create, including the path, minus any format-specific extension. format is the archive format: one of "zip" (if the zlib module is available), "tar", "gztar" (if the zlib module is available), "bztar" (if the bz2 module is available), or "xztar" (if the lzma module is available).
root_dir is a directory that will be the root directory of the archive, all paths in the archive will be relative to it; for example, we typically chdir into root_dir before creating the archive.
base_dir is the directory where we start archiving from; i.e. base_dir will be the common prefix of all files and directories in the archive. base_dir must be given relative to root_dir. See Archiving example with base_dir for how to use base_dir and root_dir together.
root_dir and base_dir both default to the current directory.
If dry_run is true, no archive is created, but the operations that would be executed are logged to logger.
owner and group are used when creating a tar archive. By default, uses the current owner and group.
logger must be an object compatible with PEP 282, usually an instance of logging.Logger.
The verbose argument is unused and deprecated.
Raises an auditing event shutil.make_archive with arguments base_name, format, root_dir, base_dir.
Note
Modifié dans la version 3.8: The modern pax (POSIX.1-2001) format is now used instead of the legacy GNU format for archives created with format="tar".
shutil.get_archive_formats()
Return a list of supported formats for archiving. Each element of the returned sequence is a tuple (name, description).
By default shutil provides these formats:
• zip: ZIP file (if the zlib module is available).
• tar: Uncompressed tar file. Uses POSIX.1-2001 pax format for new archives.
• gztar: gzip'ed tar-file (if the zlib module is available).
• bztar: bzip2'ed tar-file (if the bz2 module is available).
• xztar: xz'ed tar-file (if the lzma module is available).
You can register new formats or provide your own archiver for any existing formats, by using register_archive_format().
shutil.register_archive_format(name, function[, extra_args[, description]])
Register an archiver for the format name.
function is the callable that will be used to unpack archives. The callable will receive the base_name of the file to create, followed by the base_dir (which defaults to os.curdir) to start archiving from. Further arguments are passed as keyword arguments: owner, group, dry_run and logger (as passed in make_archive()).
If given, extra_args is a sequence of (name, value) pairs that will be used as extra keywords arguments when the archiver callable is used.
description is used by get_archive_formats() which returns the list of archivers. Defaults to an empty string.
shutil.unregister_archive_format(name)
Remove the archive format name from the list of supported formats.
shutil.unpack_archive(filename[, extract_dir[, format]])
Unpack an archive. filename is the full path of the archive.
extract_dir is the name of the target directory where the archive is unpacked. If not provided, the current working directory is used.
format is the archive format: one of "zip", "tar", "gztar", "bztar", or "xztar". Or any other format registered with register_unpack_format(). If not provided, unpack_archive() will use the archive file name extension and see if an unpacker was registered for that extension. In case none is found, a ValueError is raised.
Raises an auditing event shutil.unpack_archive with arguments filename, extract_dir, format.
Modifié dans la version 3.7: Accepts a path-like object for filename and extract_dir.
shutil.register_unpack_format(name, extensions, function[, extra_args[, description]])
Registers an unpack format. name is the name of the format and extensions is a list of extensions corresponding to the format, like .zip for Zip files.
function is the callable that will be used to unpack archives. The callable will receive the path of the archive, followed by the directory the archive must be extracted to.
When provided, extra_args is a sequence of (name, value) tuples that will be passed as keywords arguments to the callable.
description can be provided to describe the format, and will be returned by the get_unpack_formats() function.
shutil.unregister_unpack_format(name)
Unregister an unpack format. name is the name of the format.
shutil.get_unpack_formats()
Return a list of all registered formats for unpacking. Each element of the returned sequence is a tuple (name, extensions, description).
By default shutil provides these formats:
• zip: ZIP file (unpacking compressed files works only if the corresponding module is available).
• tar: uncompressed tar file.
• gztar: gzip'ed tar-file (if the zlib module is available).
• bztar: bzip2'ed tar-file (if the bz2 module is available).
• xztar: xz'ed tar-file (if the lzma module is available).
You can register new formats or provide your own unpacker for any existing formats, by using register_unpack_format().
### Archiving example¶
In this example, we create a gzip'ed tar-file archive containing all files found in the .ssh directory of the user:
>>> from shutil import make_archive
>>> import os
>>> archive_name = os.path.expanduser(os.path.join('~', 'myarchive'))
>>> root_dir = os.path.expanduser(os.path.join('~', '.ssh'))
>>> make_archive(archive_name, 'gztar', root_dir)
'/Users/tarek/myarchive.tar.gz'
The resulting archive contains:
$tar -tzvf /Users/tarek/myarchive.tar.gz drwx------ tarek/staff 0 2010-02-01 16:23:40 ./ -rw-r--r-- tarek/staff 609 2008-06-09 13:26:54 ./authorized_keys -rwxr-xr-x tarek/staff 65 2008-06-09 13:26:54 ./config -rwx------ tarek/staff 668 2008-06-09 13:26:54 ./id_dsa -rwxr-xr-x tarek/staff 609 2008-06-09 13:26:54 ./id_dsa.pub -rw------- tarek/staff 1675 2008-06-09 13:26:54 ./id_rsa -rw-r--r-- tarek/staff 397 2008-06-09 13:26:54 ./id_rsa.pub -rw-r--r-- tarek/staff 37192 2010-02-06 18:23:10 ./known_hosts ### Archiving example with base_dir¶ In this example, similar to the one above, we show how to use make_archive(), but this time with the usage of base_dir. We now have the following directory structure: $ tree tmp
tmp
└── root
└── structure
├── content
In the final archive, please_add.txt should be included, but do_not_add.txt should not. Therefore we use the following:
>>> from shutil import make_archive
>>> import os
>>> archive_name = os.path.expanduser(os.path.join('~', 'myarchive'))
>>> make_archive(
... archive_name,
... 'tar',
... root_dir='tmp/root',
... base_dir='structure/content',
... )
'/Users/tarek/my_archive.tar'
Listing the files in the resulting archive gives us:
\$ python -m tarfile -l /Users/tarek/myarchive.tar
structure/content/
## Querying the size of the output terminal¶
shutil.get_terminal_size(fallback=columns, lines)
Get the size of the terminal window.
For each of the two dimensions, the environment variable, COLUMNS and LINES respectively, is checked. If the variable is defined and the value is a positive integer, it is used.
When COLUMNS or LINES is not defined, which is the common case, the terminal connected to sys.__stdout__ is queried by invoking os.get_terminal_size().
If the terminal size cannot be successfully queried, either because the system doesn't support querying, or because we are not connected to a terminal, the value given in fallback parameter is used. fallback defaults to (80, 24) which is the default size used by many terminal emulators.
The value returned is a named tuple of type os.terminal_size.
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2021-12-07 21:35:43
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{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.30521291494369507, "perplexity": 13589.465622156922}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363418.83/warc/CC-MAIN-20211207201422-20211207231422-00105.warc.gz"}
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https://frumam.cnrs-mrs.fr/2021/02/23/seminaire-singularite/
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##### Séminaire singularité
Jeudi 25 février 14h
Abstract: The Denef-Loeser topological and motivic zeta functions are analytic invariants of holomorphic map germs $f:{\mathbb C}^n\to {\mathbb C}$, which are usually computed from embedded resolutions of $f$.
They codify some information about the topology of the Milnor fiber of the zero locus. More concretely, the Monodromy Conjecture predicts that any pole of these zeta functions is related with an eigenvalue of the monodromy at some point of $f^{-1}(0)$.
In this talk, we introduce some recent techniques that we have developed for the study of these zeta functions for $\mathbb{Q}$-divisors over orbifold varieties: a change of variables formula from relative canonical divisors, as well as a closed formula using compositions of weighted blowing-ups. Finally, we present some work in progress about applications on the study of the Monodromy Conjecture for quasi-homogeneous surface singularities.
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2021-07-26 12:40:41
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https://mathoverflow.net/questions/304282/a-cube-is-placed-inside-another-cube
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# A cube is placed inside another cube
I known following problem with two square
$$Area(1)+Area(3)=Area(2)+Area(4).$$
My question. Is this problem true for two cubes?
We place a cube $XYZT.X'Y'Z'T$ into another cube $ABCD.A'B'C'D',$ is it true that
\begin{align*} & Vol(ADD'A'.XTT'X')+Vol(BCC'B'.YZY'Z')\\ =&Vol(ABCD.XYZT)+Vol(A'B'C'D'.X'Y'Z'T')\\ =&Vol(CDD'C'.ZTT'Z')+Vol(ABB'A'XYY'X'). \end{align*}
My question. If this is true for two cubes then is this true for two similar rectangular prism?
My question. And are these problems true for two hypercubes and two similar hyperrectangles in $n$-dimesion Euclidean space?
• Why the close votes? The squares result is a neat bit of elementary geometry that I'm guessing is not (yet?) well-known (at any rate it was new to me); it's natural to ask whether it generalizes to higher dimensions, and at worst there's an easy proof of counterexample, in which case the question should be settled by giving this answer rather than closing it without comment. – Noam D. Elkies Jul 5 '18 at 2:13
• Dear Noam D. Elkies, I don't close votes this question, did any body close votes it? – Tran Quang Hung Jul 5 '18 at 2:27
• What precisely do you mean by e.g. $ADD'A'.XTT'X'$? Note that in general e.g. $A,X,X',A'$ are not coplanar. – Robert Israel Jul 5 '18 at 5:58
• Please see my figure, I mean this is volume of convex polytope $ADD'A'.XTT'X'.$ – Tran Quang Hung Jul 5 '18 at 6:00
• If it is true for hypercubes in some dimension, then it is true for similar hyperrectangles. This is just a scaling of the coordinate axes, which scales all volumes by the same factor. – jarauh Jul 5 '18 at 8:34
I believe that the answer is negative. The tidiest way to calculate is as follows. We can identify $\mathbb{R}^3$ with the purely imaginary quaternions $\mathbb{H}_0$, and let $U$ be the cube with vertices $\pm i\pm j\pm k$. Then any other cube $V$ has the form $a+zU\overline{z}$ for some $a\in\mathbb{H}_0$ and $z\in\mathbb{H}\setminus\{0\}$. For any face $F$ of $U$, let $\alpha(F)$ be the volume between $F$ and $a+zF\overline{z}$, and put $\beta(F)=\alpha(F)+\alpha(-F)$. The question is whether $\beta(F)$ is independent of $F$. To calculate $\alpha(F)$, note that we have a bijection $p$ from $[0,1]\times F$ to the relevant region given by $p(t,x)=(1-t)x+t(a+zx\overline{z})$, so we just need to integrate the Jacobian determinant of $p$ over $[0,1]\times F$. To express this determinant in terms of quaternionic algebra, note that there is an isomorphism $\Lambda^3(\mathbb{H}_0)\to\mathbb{R}$ given by $x\wedge y\wedge z\mapsto\text{Re}(xyz)$.
The derivatives of $p$ with respect to $x$ depend only on $t$. Thus, we can bring the integral over $F$ inside the determinant, which has the effect of replacing $\partial p/\partial t$ by $4$ times its value at the center $c\in F$, which is $4(a+zc\overline{z}-c)$. The derivatives of $p$ with respect to components of $x$ are of the form $m(z)+c(z)t$, so the determinant is just a quadratic in $t$. If I have everything straight, we just end up with $$\beta(F) = \frac{8}{3}(1-|z|^2)((1+|z|^2)^2 - \text{Re}(z)^2 - 3\langle z,c\rangle^2)$$ (If we call the above $\phi(z,c)$, it is not hard to check that $$\phi(z,i)+\phi(z,j)+\phi(z,k)=8(1-|z|^6) = \text{vol}(U) - \text{vol}(V),$$ as it should be.)
If $z$ is real then this is independent of $c$, so the volume formula is correct when the inner cube is parallel to the outer one. The same holds if $\langle z,c\rangle^2$ is independent of $c\in\{\pm i,\pm j,\pm k\}$; this is the case when the inner cube is obtained by rotating the outer one around one of its long axes, then shrinking and translating it. But in general the formula does not hold.
• Thank you very much Prof. Neil Strickland for your answer. I have another quesion. I hope you help me. Is it true that, sum of Volume of two tetrahedrons $$Vol(BCYZ)+Vol(A'D'X'T')=const$$ when the small cube moves inside the large. In the cases $YZ\parallel BC$ and $X'T'\parallel A'D'$ then, is it true that? $$Area(BCYZ)+Area(A'D'X'T')=const.$$ – Tran Quang Hung Jul 5 '18 at 10:46
• I don't think that the tetrahedron relation is true, and unlike the original question, I don't think that there is an interesting story behind it; the formulae are just different in an unstructured way. – Neil Strickland Jul 5 '18 at 12:54
Its also easy to give a concrete counter-example: Take the cube with vertices $$\left(\frac{273}{340},\,\frac{79}{68},\,\frac{13}{20}\right) , \left(\frac{407}{340},\,\frac{57}{68},\,\frac{27}{20}\right) , \left(\frac{239}{340},\,\frac{789}{884},\,\frac{337}{260}\right) , \left(\frac{249}{340},\,\frac{621}{884},\,\frac{217}{260}\right) , \left(\frac{263}{340},\,\frac{1195}{884},\,\frac{289}{260}\right) , \left(\frac{417}{340},\,\frac{573}{884},\,\frac{231}{260}\right) , \left(\frac{431}{340},\,\frac{1147}{884},\,\frac{303}{260}\right) , \left(\frac{441}{340},\,\frac{979}{884},\,\frac{183}{260}\right)$$ inside a $0-2$-cube. Then the three volumes you ask for are $\frac{24421}{8840}\neq \frac{24563}{8840}\neq \frac{24491}{8840}$. Here is a picture of the configuration:
The outer cube has volume 8 and the inner cube has volume $1/8$. As a sanity check, lets see if everything adds up: $$\frac{24421}{8840} + \frac{24563}{8840} + \frac{24491}{8840} = \frac{14695}{1768}$$ and $$\frac{14695}{1768} + \frac{1}{8} -8 =\frac{193}{442}.$$
Since the edges of the outer cube are not coplanar with the edges, we have now double counted all the 12 tetrahedra, that arise from the fact that these pairs of edges are non-coplanar. Their volumes are $$\frac{471}{8840} , \frac{11}{170} , \frac{59}{2210} , \frac{29}{680} , \frac{49}{1768} , \frac{29}{8840} , \frac{29}{680} , \frac{29}{8840} , \frac{59}{2210} , \frac{49}{1768} , \frac{11}{170} , \frac{471}{8840}$$ and sure enough the sum of these numbers is $\frac{193}{442}$. Here is a picture of all those tetrahedra:
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2020-06-04 21:48:01
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https://socratic.org/questions/what-are-the-critical-points-of-f-x-2x-3-3x-2-12x
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# What are the critical points of f(x) = 2x^3-3x^2-12x?
Oct 28, 2015
The critical points are $- 1$ and $2$.
$f ' \left(x\right) = 6 {x}^{2} - 6 x - 12 = 6 \left({x}^{2} - x - 2\right)$
So. to find its zeros, we must look for the zeroes of $\left({x}^{2} - x - 2\right)$, which are $- 1$ and $2$. So, these are the critical points of $f \left(x\right)$.
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2019-06-16 23:06:41
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https://www.popflock.com/learn?s=Response_factor
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Response Factor
Get Response Factor essential facts below. View Videos or join the Response Factor discussion. Add Response Factor to your PopFlock.com topic list for future reference or share this resource on social media.
Response Factor
Response factor, usually in chromatography and spectroscopy, is the ratio between a signal produced by an analyte, and the quantity of analyte which produces the signal. Ideally, and for easy computation, this ratio is unity (one). In real-world scenarios, this is often not the case.
## Expression
The response factor ${\displaystyle f_{i}}$ can be expressed on a molar, volume or mass[1] basis. Where the true amount of sample and standard are equal:
${\displaystyle f_{i}={\frac {A_{i}}{A_{st}}}f_{st}}$
where A is the signal (e.g. peak area) and the subscript i indicates the sample and the subscript st indicates the standard.[2] The response factor of the standard is assigned an arbitrary factor, for example 1 or 100. Response factor of sample/Response factor of standard=RRF
## Chromatography
One of the main reasons to use response factors is to compensate for the irreproducibility of manual injections into a gas chromatograph (GC). Injection volumes for GCs can be 1 microliter (µL) or less and are difficult to reproduce. Differences in the volume of injected analyte leads to differences in the areas of the peaks in the chromatogram and any quantitative results are suspect.
To compensate for this error, a known amount of an internal standard (a second compound that does not interfere with the analysis of the primary analyte) is added to all solutions (standards and unknowns). This way if the injection volumes (and hence the peak areas) differ slightly, the ratio of the areas of the analyte and the internal standard will remain constant from one run to the next.
This comparison of runs also applies to solutions with different concentrations of the analyte. The area of the internal standard becomes the value to which all other areas are referenced. Below is the mathematical derivation and application of this method.
Consider an analysis of octane (C8H18) using nonane (C9H20) as the internal standard. The 3 chromatograms below are for 3 different samples.
The amount of octane in each sample is different, but the amount of nonane is the same (in practice this is not a requirement). Due to scaling, the areas of the nonane peak appear to have different areas, but in reality the areas are identical. Therefore, the relative amounts of octane in each sample increases in the order of mixture 1 (least) < mixture 3 < mixture 2 (most).
This conclusion is reached because the ratio of the area of octane to that of nonane is the least in mixture 1 and the most, in mixture 2. Mixture 3 has an intermediate ratio. This ratio can be written as ${\displaystyle Area_{octane}/Area_{nonane}}$.
In chromatography, the area of a peak is proportional to the number of moles (n) times some constant of proportionality (k), Area = k×n. The number of moles of compound is equal to the concentration (molarity, M) times the volume, n = MV. From these equations, the following derivation is made:
${\displaystyle {{Area_{octane}} \over {Area_{nonane}}}={{k_{octane}\times M_{octane}\times V_{octane}} \over {k_{nonane}\times M_{nonane}\times V_{nonane}}}}$
Since both compounds are in the same solution and are injected together, the volume terms are equal and cancel out. The above equation is then rearranged to solve for the ratio of the k's. This ratio is then called the response factor, F.
${\displaystyle F={k_{octane} \over k_{nonane}}={{Area_{octane}/M_{octane}} \over {Area_{nonane}/M_{nonane}}}}$
The response factor, F, is equal to the ratios of the k's, which are constant. Therefore, F is constant. What this means is that regardless of the amounts of octane and nonane in solution, the ratio of the ratios of area to concentration will always yield a constant.
In practice, a solution containing known amounts of both octane and nonane is injected into a GC and a response factor, F, is calculated. Then a separate solution with an unknown amount of octane and a known amount of nonane is injected. The response factor is applied to the data from the second solution and the unknown concentration of the octane is found.
${\displaystyle {\left({{Area_{octane}/M_{octane}} \over {Area_{nonane}/M_{nonane}}}\right)}_{1}=F={\left({{Area_{octane}/M_{octane}} \over {Area_{nonane}/M_{nonane}}}\right)}_{2}}$
This example deals with the analysis of octane and nonane, but can be applied to any two compounds.
## References
1. ^ Ramus TL; Hein SJ; Thomas LC (August 1987). "Determinations of polychlorinated biphenyl isomers by response factor calibration". J. Chromatogr. 404 (1): 155-62. doi:10.1016/S0021-9673(01)86846-1. PMID 3119645.
2. ^ Orange Book. Compendium of Analytical Nomenclature (PDF). IUPAC.
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2021-07-28 10:05:21
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https://philpeople.org/profiles/edward-eugene-kleist/publications
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• 17
##### Phenomenology’s Constitutive Paradox Idealistic Studies 48 (2): 133-147. 2018.
I provide a phenomenological response to Quentin Meillassoux’s “realist” criticism of phenomenology and I explore the resources and limits of phenomenology in its own attempt to grapple with the paradox Meillassoux believes sinks it: subjectivity has priority over the physical reality it constitutes despite the anteriority and posteriority of that physical reality to subjectivity. I first offer a corrective to Meillassoux’s interpretation of Husserl. Then, I turn to Merleau-Ponty’s lectures on t…Read more
• 15
• 7
##### The Freedom to Design Nature: Kant's Strong Ought→Can Inference in 21st Century Perspective Cosmos and History : The Journal of Natural and Social Philosophy 1 (2): 213-221. 2005.
Kant’s attempts to formulate a conception of the harmony of nature and freedom have two logical presuppositions. The first presupposition is separation of ought and is, which provides a logical formulation of the separation of freedom and nature. Kant might well have settled on the view that the separation between nature and freedom cannot be bridged. Why did Kant attempt to overcome said separation? The second presupposition of Kant’s project to bridge nature and freedom involves an ought→can i…Read more
• ##### Contemplation and Judgment in Kant's Aesthetics Dissertation, Boston College. 1994.
The Critique of Judgment aims to account for the affective sharing of a common world of appearance. To accomplish this project, Kant retrieves a connection between contemplation and judgment which had lain dormant in the philosophical tradition since the time of Plato. Kant rescues the theme of contemplatio or $\theta\varepsilon\omega\rho\acute\iota\alpha$ from the Neo-platonist tradition culminating in Leibniz and Shaftesbury. This tradition took beauty as the motivation for an intuitive assimi…Read more
• 22
• 2
##### Imagination as a reflection of value-commitment Ethic@ - An International Journal for Moral Philosophy 6 (2): 177-187. 2007.
Hume remarked on how our moral value-commitments set limits for what we are willing to imagine. Moral values also guide imagination when we envision variant scenarios and options for action. How do values reveal themselves through imagining? What does the manner through which valuesappear tell us about the nature of values? Imagination furnishes a non-perceptual manner of arriving at moral determinations anchored to the irreducibly first-person experience of moral approval and disapproval. The c…Read more
• 95
##### Schopenhauer on the Individuation and Teleology of Intelligible Character Idealistic Studies 40 (1-2): 15-26. 2010.
A problem arises in Schopenhauer’s claim that each individual person’s will, or intelligible character, is timeless. The principium individuationis depends upon spatio-temporal determinations governing the world as representation. As individual, one’s individual character would seem to depend upon spatio-temporalconditions. Yet, Schopenhauer adopts the Kantian distinction between empirical character and intelligible character, with the individual’s intelligible characterremaining the timeless Di…Read more
• 23
##### The Freedom to Design Nature: Kant's Strong Ought→ Can Inference in 21st Century Perspective Cosmos and History 1 (2): 213-221. 2005.
Kant’s attempts to formulate a conception of the harmony of nature and freedom have two logical presuppositions. The first presupposition is separation of ought and is, which provides a logical formulation of the separation of freedom and nature. Kant might well have settled on the view that the separation between nature and freedom cannot be bridged. Why did Kant attempt to overcome said separation? The second presupposition of Kant’s project to bridge nature and freedom involves an ought→can i…Read more
• 5
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2020-09-28 09:18:55
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https://codereview.stackexchange.com/questions/129184/marshaling-data-for-a-game
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# Marshaling data for a game
I've been working on a game that I used to play as a kid. This game was disassembled then converted to C# by someone else and they hosted it on Github. I've forked it and started working on it some. One of the things I've been trying to convert over is how to read back the files that were saved during a save operation of the game.
Now, although I'm only asking about 1 specific field in this structure, having any tips on how to clean up my code would be much appreciated. After the first field the next 4 fields I have yet to convert into a meaningful structure so I've left them as large chunks of data.
Here is the structure I'm marshling the data into:
[StructLayout(LayoutKind.Sequential, CharSet = CharSet.Ansi, Pack = 1, Size = 0x335D)]
public struct CurseSaveGame
{
public byte GameArea;
[MarshalAs(UnmanagedType.ByValArray, SizeConst = 0x800)]
public byte[] AreaPointer1;
[MarshalAs(UnmanagedType.ByValArray, SizeConst = 0x800)]
public byte[] AreaPointer2;
[MarshalAs(UnmanagedType.ByValArray, SizeConst = 0x200)]
public ushort[] SomeStructure;
[MarshalAs(UnmanagedType.ByValArray, SizeConst = 0x1E00)]
public byte[] EclPointer;
public sbyte MapPosX;
public sbyte MapPosY;
public byte MapDirection;
public byte MapWallType;
public byte MapWallRoof;
public GameState LastGameState;
public GameState GameState;
public SetBlock SetBlock1;
public SetBlock SetBlock2;
public SetBlock SetBlock3;
public byte[] ToByteArray() => Extensions.ToByteArray(this);
public static CurseSaveGame Parse(byte[] array) => Extensions.MarshalAs<CurseSaveGame>(array);
}
Here are the structures that I've mapped over thus far in that structure
[StructLayout(LayoutKind.Sequential, CharSet = CharSet.Ansi, Pack = 1, Size = 4)]
public struct SetBlock
{
public short BlockId;
public short SetId;
}
[StructLayout(LayoutKind.Sequential, CharSet = CharSet.Ansi, Pack = 1, Size = 0x149)]
{
public byte NumberOfPlayersInParty;
}
[StructLayout(LayoutKind.Sequential, CharSet = CharSet.Ansi, Pack = 1, Size = 0x29)]
{
private byte FileNameLength;
[MarshalAs(UnmanagedType.ByValTStr, SizeConst = 0x08)]
public string FileName;
[MarshalAs(UnmanagedType.ByValArray, SizeConst = 0x20)]
private byte[] Houses;
}
public enum GameState : byte
{
Shop = 1,
Camping = 2,
WildernessMap = 3,
DungeonMap = 4,
Combat = 5,
AfterCombat = 6,
EndGame = 7
}
The part in question is the structure PlayerLoadFiles. I think that it would be best to have it more like:
[StructLayout(LayoutKind.Sequential, CharSet = CharSet.Ansi, Pack = 1, Size = 0x149)]
{
public byte NumberOfPlayersInParty;
}
but when I do that (with different attributes attached to the PlayerLoadFiles field I keep getting the same error message when I Parse the array. Type 'CurseSaveGame' cannot be marshaled as an unmanaged structure; no meaningful size or offset can be computed. Mentally I feel like this is the closest attribute for LoadFiles but again it fails:
[MarshalAs(UnmanagedType.CustomMarshaler, MarshalTypeRef = typeof(CharacterLoadFile), SizeConst = 8)]
I suppose I can just keep the 8 structures as it works and I have the number of players. I suppose I could put a public getter on there with all 8 players as an array then use Linq's Take method to only take the correct number of players.
For testing purposes I did the following. I did a hex dump of the last 0x149 bytes and did a hex string to byte array, then marshaled that as the PlayerLoadFiles. I do it in Linqpad so it is fast and easy for me to visualize.
var loadData = new List<byte[]>
{
@"06-08-43-48-52-44-41-54-41-31-00-00-00-0A-00-14-00 0C-00-AC-01-0D-0A-F3-00-08-01-C0-18-B2-21-C4-3C B8-02-20-00-00-00-00-00-18-08-43-48-52-44-41-54 41-32-46-00-46-00-40-00-96-13-FF-00-20-04-DA-3C 92-03-A5-47-01-00-20-00-04-00-00-00-00-00-18-00 28-00-08-43-48-52-44-41-54-41-33-3D-02-00-06-01 02-01-03-05-06-00-00-00-00-00-00-00-00-00-00-00 00-00-00-00-00-00-00-00-00-00-00-08-43-48-52-44 41-54-41-34-00-00-00-00-00-1E-06-01-06-02-4E-B6 54-77-00-84-54-77-00-66-54-77-00-06-59-65-73-20 4E-6F-19-00-08-43-48-52-44-41-54-41-35-09-09-0B 0B-0D-0D-0F-0F-11-11-13-13-15-15-17-17-19-19-00 00-00-00-00-1E-47-61-6D-65-20-6E-6F-74-08-43-48 52-44-41-54-41-36-51-75-69-74-20-61-6E-79-77-61 79-3F-20-20-4D-20-54-20-48-20-56-20-41-4E-C4-3D 31-12-A5-47-99-3D-B2-21-00-00-00-00-0F-00-0A-00 42-12-A5-47-C4-11-A5-47-4E-00-20-06-59-65-73-20 4E-6F-4E-00-1E-47-61-6D-65-20-6E-6F-74-20-73-61 76-65-64-2E-20-20-51-75-69-74-20-61-6E-79-77-61 79-3F-20-00-00-00-18-00-28-00-CE-3D-02-4E-F8-3E 47-06-82-38-0F-5E-0A-5E".ToByteArray(),
};
public static T MarshalAs<T>(byte[] rawDataStructure) where T : struct
{
var type = typeof(T);
int size = Marshal.SizeOf(type);
IntPtr ptr = Marshal.AllocHGlobal(size);
Marshal.Copy(rawDataStructure, 0, ptr, size);
T structure = (T)Marshal.PtrToStructure(ptr, type);
Marshal.FreeHGlobal(ptr);
return structure;
}
public static byte[] ToByteArray(this string hexString)
{
hexString = hexString.ToUpper();
hexString = System.Text.RegularExpressions.Regex.Replace(hexString, "[^0-9A-F.]", "");
if (hexString.Length % 2 == 1)
throw new Exception("The binary key cannot have an odd number of digits");
byte[] arr = new byte[hexString.Length >> 1];
for (int i = 0; i < hexString.Length >> 1; ++i)
{
arr[i] = (byte)((GetHexVal(hexString[i << 1]) << 4) + (GetHexVal(hexString[(i << 1) + 1])));
}
return arr;
}
private static int GetHexVal(char hex)
{
int val = (int)hex;
return val - (val < 58 ? 48 : 55);
}
Any feedback on how I can shrink this up some more and make it more readable would be great. Although I'm not new to byte level data this marshaling thing is sorta new to me like this.
Note: this is for the game Curse of the Azure bonds. You can see my fork of the code here on GitHub.
If there really always is 8 player objects, even in the cases where NumberOfPlayersInParty is less than 8, then you can use ByValArray:
[MarshalAs(UnmanagedType.ByValArray, SizeConst = 0x08)]
But if there is only as many player object as NumberOfPlayersInParty indicates, then even your original solution is not correct and you will probably have to resort to deserializing the data at least partially manually.
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2019-11-20 21:07:00
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